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Recent questions and answers in Operating System

1 vote
3 answers
2
UGCNET-Jan2017-III: 53
One of the disadvantages of user level threads compared to Kernel level thread is If a user level thread of a process executes a system call, all threads in that process are blocked. Scheduling is application dependent. Thread switching doesn’t require kernel mode privileges. The library procedures invoked for thread management in user level threads are local procedures.
One of the disadvantages of user level threads compared to Kernel level thread is If a user level thread of a process executes a system call, all threads in that process are blocked. Scheduling is application dependent. Thread switching doesn’t require kernel mode privileges. The library procedures invoked for thread management in user level threads are local procedures.
answered Jan 14 in Operating System anjli 270 views
0 votes
3 answers
3
UGCNET-Jan2017-III: 51
Match the following for Unix file system : ... b-i, c-ii, d-iv a-iii, b-i, c-iv, d-ii a-iv, b-iii, c-ii, d-i a-iv, b-iii, c-i, d-ii
Match the following for Unix file system : ... Codes: a-iii, b-i, c-ii, d-iv a-iii, b-i, c-iv, d-ii a-iv, b-iii, c-ii, d-i a-iv, b-iii, c-i, d-ii
answered Jan 14 in Operating System anjli 204 views
0 votes
2 answers
4
NIELIT 2017 OCT Scientific Assistant A (IT) - Section B: 23
Which two are valid constructions for Thread? Thread(Runnable r, String name) Thread() Thread(int priority) Thread(Runnable r, ThreadGroup g) Thread(Runnable r, int priority) $1$ and $3$ $2$ and $4$ $1$ and $2$ $2$ and $5$
Which two are valid constructions for Thread? Thread(Runnable r, String name) Thread() Thread(int priority) Thread(Runnable r, ThreadGroup g) Thread(Runnable r, int priority) $1$ and $3$ $2$ and $4$ $1$ and $2$ $2$ and $5$
answered Jan 14 in Operating System anjli 215 views
22 votes
3 answers
5
GATE2017-1-24
Consider the following CPU processes with arrival times (in milliseconds) and length of CPU bursts (in milliseconds) as given below: ... time first scheduling algorithm is used to schedule the processes, then the average waiting time across all processes is _____________ milliseconds.
Consider the following CPU processes with arrival times (in milliseconds) and length of CPU bursts (in milliseconds) as given below: ... remaining time first scheduling algorithm is used to schedule the processes, then the average waiting time across all processes is _____________ milliseconds.
answered Jan 14 in Operating System anjli 4.5k views
0 votes
2 answers
6
Process Synchronisation ( binary semaphores)
Consider the below program as two concurrent processes: Semaphore x:= 0; P1: repeat forever P2: repeat forever 1.V(x); 4. P(x); 2.Compute; 5.Compute; 3.P(x); 6.V(x); Consider the following statements about process P1 & P2: (1) ... chance to run , and it wil starve (THIS IS WHAT I THOUGHT THE ANSWER SHOULD BE , PLEASE COMMENT, IF I AM WRONG ANYWHERE)
Consider the below program as two concurrent processes: Semaphore x:= 0; P1: repeat forever P2: repeat forever 1.V(x); 4. P(x); 2.Compute; 5.Compute; 3.P(x); 6.V(x); Consider the following statements about process P1 & P2: (1) It is possible for process P1 to ... never get the chance to run , and it wil starve (THIS IS WHAT I THOUGHT THE ANSWER SHOULD BE , PLEASE COMMENT, IF I AM WRONG ANYWHERE)
answered Jan 14 in Operating System anjli 379 views
2 votes
3 answers
8
Scheduling
Consider the processes P1, P2, P3, P4 whose arrival times are 0, 1, 2, 3 and Burst times are 5, 2, 13, 7. If the Context Switching time is 1 unit (ms), what is the average waiting time if Shortest Job Next scheduling algorithm is used? (a) 7.5 (b) 6.25 (c) 6.5 (d) None (The Doubt i am facing here is , would we be covering the context switching time in the starting ?)
Consider the processes P1, P2, P3, P4 whose arrival times are 0, 1, 2, 3 and Burst times are 5, 2, 13, 7. If the Context Switching time is 1 unit (ms), what is the average waiting time if Shortest Job Next scheduling algorithm is used? (a) 7.5 (b) 6.25 (c) 6.5 (d) None (The Doubt i am facing here is , would we be covering the context switching time in the starting ?)
answered Jan 14 in Operating System anjli 1.5k views
1 vote
3 answers
9
UGCNET-July-2018-II: 59
Consider the following three processes with the arrival time and CPU burst time given in miliseconds: Process Arrival Time Burst Time $P_1$ 0 7 $P_2$ 1 4 $P_3$ 2 8 The Gantt Chart for preemptive SJF scheduling algorithm is _______
Consider the following three processes with the arrival time and CPU burst time given in miliseconds: Process Arrival Time Burst Time $P_1$ 0 7 $P_2$ 1 4 $P_3$ 2 8 The Gantt Chart for preemptive SJF scheduling algorithm is _______
answered Jan 14 in Operating System anjli 1k views
73 votes
14 answers
11
GATE2015-1-46
Consider a uniprocessor system executing three tasks $T_{1}, T_{2}$ and $T_{3}$ each of which is composed of an infinite sequence of jobs (or instances) which arrive periodically at intervals of $3$, $7$ and $20$ milliseconds, ... the 1st millisecond and task preemptions are allowed, the first instance of $T_{3}$ completes its execution at the end of_____________________milliseconds.
Consider a uniprocessor system executing three tasks $T_{1}, T_{2}$ and $T_{3}$ each of which is composed of an infinite sequence of jobs (or instances) which arrive periodically at intervals of $3$, $7$ and $20$ milliseconds, respectively ... the 1st millisecond and task preemptions are allowed, the first instance of $T_{3}$ completes its execution at the end of_____________________milliseconds.
answered Jan 14 in Operating System tusharSingh 18.9k views
0 votes
2 answers
12
NIELIT 2017 July Scientist B (IT) - Section B: 48
A scheduling algorithm assigns priority proportional to the waiting time of a process. Every process starts with priority zero (the lowest priority). The scheduler re-evaluates the process priorities every $T$ time units and ... to the shortest-job-first algorithm. This algorithm is equivalent to the shortest-remaining time-first algorithm.
A scheduling algorithm assigns priority proportional to the waiting time of a process. Every process starts with priority zero (the lowest priority). The scheduler re-evaluates the process priorities every $T$ time units and decides the next process to ... algorithm is equivalent to the shortest-job-first algorithm. This algorithm is equivalent to the shortest-remaining time-first algorithm.
answered Jan 13 in Operating System anjli 374 views
56 votes
5 answers
14
GATE2003-79
A processor uses $\text{2-level}$ page tables for virtual to physical address translation. Page tables for both levels are stored in the main memory. Virtual and physical addresses are both $32$ bits wide. The memory is byte addressable. For virtual to physical address translation, the $10$ ... the page tables of this process is $\text{8 KB}$ $\text{12 KB}$ $\text{16 KB}$ $\text{20 KB}$
A processor uses $\text{2-level}$ page tables for virtual to physical address translation. Page tables for both levels are stored in the main memory. Virtual and physical addresses are both $32$ bits wide. The memory is byte addressable. For virtual to physical address translation, the $10$ most ... the page tables of this process is $\text{8 KB}$ $\text{12 KB}$ $\text{16 KB}$ $\text{20 KB}$
answered Jan 13 in Operating System tusharSingh 12.5k views
3 votes
2 answers
15
Process Synchronization
Mutual Exclusion is important for deadlock to happen. Is this statement necessarily true always ?
Mutual Exclusion is important for deadlock to happen. Is this statement necessarily true always ?
answered Jan 11 in Operating System Sneha negi 853 views
0 votes
2 answers
16
UGCNET-Oct2020-II: 65
Match $\text{List I}$ with $\text{List II}$ ... D-IV$ $A-II, B-IV, C-III, D-I$ $A-II, B-IV, C-I, D-III$ $A-IV, B-III, C-II, D-I$ Match $
Match $\text{List I}$ with $\text{List II}$ ... C-III, D-IV$ $A-II, B-IV, C-III, D-I$ $A-II, B-IV, C-I, D-III$ $A-IV, B-III, C-II, D-I$ Match $
answered Jan 10 in Operating System eshita1997 79 views
30 votes
5 answers
18
GATE2009-52
A hard disk has $63$ sectors per track, $10$ platters each with $2$ recording surfaces and $1000$ cylinders. The address of a sector is given as a triple $\langle c, h, s \rangle$, where $c$ is the cylinder number, $h$ is the surface number and $s$ is the sector number. ... $\langle 0, 15, 31 \rangle$ $\langle 0, 16, 30 \rangle$ $\langle 0, 16, 31 \rangle$ $\langle 0, 17, 31 \rangle$
A hard disk has $63$ sectors per track, $10$ platters each with $2$ recording surfaces and $1000$ cylinders. The address of a sector is given as a triple $\langle c, h, s \rangle$, where $c$ is the cylinder number, $h$ is the surface number and $s$ is the sector number. Thus, the ... is $\langle 0, 15, 31 \rangle$ $\langle 0, 16, 30 \rangle$ $\langle 0, 16, 31 \rangle$ $\langle 0, 17, 31 \rangle$
answered Jan 10 in Operating System eshita1997 5.3k views
27 votes
3 answers
19
GATE2005-20
Normally user programs are prevented from handling I/O directly by I/O instructions in them. For CPUs having explicit I/O instructions, such I/O protection is ensured by having the I/O instruction privileged. In a CPU with memory mapped I/O, there is ... s) I/O protection is ensured by a hardware trap I/O protection is ensured during system configuration I/O protection is not possible
Normally user programs are prevented from handling I/O directly by I/O instructions in them. For CPUs having explicit I/O instructions, such I/O protection is ensured by having the I/O instruction privileged. In a CPU with memory mapped I/O, there is no explicit ... (s) I/O protection is ensured by a hardware trap I/O protection is ensured during system configuration I/O protection is not possible
answered Jan 8 in Operating System PSDesai09 6.7k views
0 votes
2 answers
20
Andrew S. Tanenbaum (OS) Edition 4 Exercise 6 Question 31 (Page No. 468)
One way to prevent deadlocks is to eliminate the hold-and-wait condition. In the text it was proposed that before asking for a new resource, a process must first release whatever resources it already holds ( ... new resource but lose some of the existing ones to competing processes. Propose an improvement to this scheme.
One way to prevent deadlocks is to eliminate the hold-and-wait condition. In the text it was proposed that before asking for a new resource, a process must first release whatever resources it already holds (assuming that is possible). However, doing so introduces ... that it may get the new resource but lose some of the existing ones to competing processes. Propose an improvement to this scheme.
answered Jan 6 in Operating System Joey 108 views
3 votes
2 answers
21
Process Synchronisation
P0: P(S), P(Q), Print(“Hello”), V(Q), V(S) P1: P(Q), P(S), Print(“Hi”), V(Q), V(S) Where S & Q are two semaphores initialized to 1. In the above situation: (a) Deadlock may occur (b) Bounded Waiting is satisfied (c) Deadlock never occurs d) Both (a)&(b)
P0: P(S), P(Q), Print(“Hello”), V(Q), V(S) P1: P(Q), P(S), Print(“Hi”), V(Q), V(S) Where S & Q are two semaphores initialized to 1. In the above situation: (a) Deadlock may occur (b) Bounded Waiting is satisfied (c) Deadlock never occurs d) Both (a)&(b)
answered Jan 3 in Operating System pawanchoure 568 views
0 votes
1 answer
22
MADE EASY OS
Which of the following allocation scheme does not support direct access in file allocation scheme? (a)Linked Allocation (b)Indexed Allocation (c)Contiguous Allocation both (a) and (b)
Which of the following allocation scheme does not support direct access in file allocation scheme? (a)Linked Allocation (b)Indexed Allocation (c)Contiguous Allocation both (a) and (b)
answered Dec 31, 2020 in Operating System Abhi Prajapati 585 views
1 vote
1 answer
23
operating system
Suppose the following jobs are to be executed in a uniprocessor system. Assume the overhead of context switching is one time unit. The average process turnaround time, thenormalized turnaround time for process 2, and the processor efficiency using SRTF is, a. 11.4, 2, 83.3% b. ... .625, 83.3% d. 11.6, 3.75, 80.6% Does we have to count the context switch for the first process also?
Suppose the following jobs are to be executed in a uniprocessor system. Assume the overhead of context switching is one time unit. The average process turnaround time, thenormalized turnaround time for process 2, and the processor efficiency using SRTF is, a. 11.4, 2, 83.3% b. 18.6, 3.375 ... .8, 1.625, 83.3% d. 11.6, 3.75, 80.6% Does we have to count the context switch for the first process also?
answered Dec 28, 2020 in Operating System prakhar2810 467 views
42 votes
7 answers
24
GATE2013-34
A shared variable $x$, initialized to zero, is operated on by four concurrent processes $W, X, Y, Z$ as follows. Each of the processes $W$ and $X$ reads $x$ from memory, increments by one, stores it to memory, and then terminates. Each of the processes $Y$ and ... $x$ after all processes complete execution? $-2$ $-1$ $1$ $2$
A shared variable $x$, initialized to zero, is operated on by four concurrent processes $W, X, Y, Z$ as follows. Each of the processes $W$ and $X$ reads $x$ from memory, increments by one, stores it to memory, and then terminates. Each of the processes $Y$ and $Z$ ... $S$ is initialized to two. What is the maximum possible value of $x$ after all processes complete execution? $-2$ $-1$ $1$ $2$
answered Dec 25, 2020 in Operating System StoneHeart 10.5k views
22 votes
5 answers
25
GATE2002-1.22
Which of the following scheduling algorithms is non-preemptive? Round Robin First-In First-Out Multilevel Queue Scheduling Multilevel Queue Scheduling with Feedback
Which of the following scheduling algorithms is non-preemptive? Round Robin First-In First-Out Multilevel Queue Scheduling Multilevel Queue Scheduling with Feedback
answered Dec 25, 2020 in Operating System StoneHeart 4k views
0 votes
1 answer
26
UGCNET-Oct2020-II: 15
Consider a hypothetical machine with $3$ pages of physical memory, $5$ pages of virtual memory, and $<A, B, C, D, A, B, E, A, B, C, D, E, B, A, B>$ as the stream of page reference by an application. If $P$ and $Q$ are the number of page faults that ... , then $(P,Q)=$ _______(Assuming enough space for storing $3$ page frames) $(11,10)$ $(12,11)$ $(10,11)$ $(11,12)$
Consider a hypothetical machine with $3$ pages of physical memory, $5$ pages of virtual memory, and $<A, B, C, D, A, B, E, A, B, C, D, E, B, A, B>$ as the stream of page reference by an application. If $P$ and $Q$ are the number of page faults that the application ... respectively, then $(P,Q)=$ _______(Assuming enough space for storing $3$ page frames) $(11,10)$ $(12,11)$ $(10,11)$ $(11,12)$
asked Nov 20, 2020 in Operating System jothee 160 views
0 votes
2 answers
27
UGCNET-Oct2020-II: 16
Consider a disk system having $60$ cylinders. Disk requests are received by a disk drive for cylinders $10,22,20,2,40,6$ and $38$, in that order. Assuming the disk head is currently at cylinder $20$, what is the time taken to ... adjacent one and Shortest Seek Time First (SSTF) algorithm is used? $240$ milliseconds $96$ milliseconds $120$ milliseconds $112$ milliseconds
Consider a disk system having $60$ cylinders. Disk requests are received by a disk drive for cylinders $10,22,20,2,40,6$ and $38$, in that order. Assuming the disk head is currently at cylinder $20$, what is the time taken to satisfy all the ... cylinder to adjacent one and Shortest Seek Time First (SSTF) algorithm is used? $240$ milliseconds $96$ milliseconds $120$ milliseconds $112$ milliseconds
asked Nov 20, 2020 in Operating System jothee 95 views
0 votes
0 answers
28
UGCNET-Oct2020-II: 17
Suppose you have a Linux file system where the block size is $2K$ bytes, a disk address is $32$ bits, and an $i-$node contains the disk addresses of the first $12$ direct blocks of file, a single indirect block and a double indirect block. Approximately, what is the largest file that can be represented by an $i-$node? $513$ Kbytes $513$ MBytes $537$ Mbytes $537$ KBytes
Suppose you have a Linux file system where the block size is $2K$ bytes, a disk address is $32$ bits, and an $i-$node contains the disk addresses of the first $12$ direct blocks of file, a single indirect block and a double indirect block. Approximately, what is the largest file that can be represented by an $i-$node? $513$ Kbytes $513$ MBytes $537$ Mbytes $537$ KBytes
asked Nov 20, 2020 in Operating System jothee 65 views
2 votes
2 answers
29
UGCNET-Oct2020-II: 18
Consider a single-level page table system, with the page table stored in the memory. If the hit rate to TLB is $80\%$, and it takes $15$ nanoseconds to search the $TLB$, and $150$ nanoseconds to access the main memory, then what is the effective memory access time, in nanoseconds? $185$ $195$ $205$ $175$
Consider a single-level page table system, with the page table stored in the memory. If the hit rate to TLB is $80\%$, and it takes $15$ nanoseconds to search the $TLB$, and $150$ nanoseconds to access the main memory, then what is the effective memory access time, in nanoseconds? $185$ $195$ $205$ $175$
asked Nov 20, 2020 in Operating System jothee 87 views
1 vote
1 answer
30
UGCNET-Oct2020-II: 84
Assuming that the system call $\text{fork}()$ never fails, consider the following C program $P1$ and $P2$ ... $II$ are false Statement $I$ is correct but Statement $II$ is false Statement $I$ is incorrect but Statement $II$ is true
Assuming that the system call $\text{fork}()$ never fails, consider the following C program $P1$ and $P2$ ... Both Statement $I$ and Statement $II$ are false Statement $I$ is correct but Statement $II$ is false Statement $I$ is incorrect but Statement $II$ is true
asked Nov 20, 2020 in Operating System jothee 57 views
0 votes
1 answer
31
UGCNET-Oct2020-II: 91
Comprehension: For the question given below: concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds. If $T$ is the capacity of a track in bytes, and $S$ is the capacity of ... $(50 K, 50000 K)$ $(25 K, 25000 K)$ $(25 K, 50000 K)$ $(40 K, 36000 K)$
Comprehension: For the question given below: concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds. If $T$ is the capacity of a track in bytes, and $S$ is the capacity of each surface in ... $(T,S)=$ _______ $(50 K, 50000 K)$ $(25 K, 25000 K)$ $(25 K, 50000 K)$ $(40 K, 36000 K)$
asked Nov 20, 2020 in Operating System jothee 42 views
0 votes
1 answer
32
UGCNET-Oct2020-II: 92
Comprehension: For the question given below: concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds. What is the capacity of the disk, in bytes? $25,000 K$ $500,000 K$ $250,000 K$ $50,000 K$
Comprehension: For the question given below: concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds. What is the capacity of the disk, in bytes? $25,000 K$ $500,000 K$ $250,000 K$ $50,000 K$
asked Nov 20, 2020 in Operating System jothee 38 views
0 votes
1 answer
33
UGCNET-Oct2020-II: 93
Comprehension: For the question given below: concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds. Given below are two statements: ... are false Statement $I$ is correct but Statement $II$ is false Statement $I$ is incorrect but Statement $II$ is true
Comprehension: For the question given below: concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds. Given below are two statements: Statement $I$: The ... $II$ are false Statement $I$ is correct but Statement $II$ is false Statement $I$ is incorrect but Statement $II$ is true
asked Nov 20, 2020 in Operating System jothee 31 views
0 votes
1 answer
34
UGCNET-Oct2020-II: 94
Comprehension: For the question given below, concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ ... , then approximately what is the maximum rotational delay? $0.011$ seconds $0.11$ seconds $0.0011$ seconds $1.1$ seconds
Comprehension: For the question given below, concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds. If the disk platters rotate at $5400$ rpm ( ... minute), then approximately what is the maximum rotational delay? $0.011$ seconds $0.11$ seconds $0.0011$ seconds $1.1$ seconds
asked Nov 20, 2020 in Operating System jothee 30 views
0 votes
1 answer
35
UGCNET-Oct2020-II: 95
For the question given below, concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds. If one track of data can be transferred per revolution, ... is the data transfer rate? $2,850$ KBytes/second $4,500$ KBytes/second $5,700$ KBytes/second $2,250$ KBytes/second
For the question given below, concern a disk with a sector size of $512$ bytes, $2000$ tracks per surface, $50$ sectors per track, five double-sided platters, and average seek time of $10$ milliseconds. If one track of data can be transferred per revolution, then what is the data transfer rate? $2,850$ KBytes/second $4,500$ KBytes/second $5,700$ KBytes/second $2,250$ KBytes/second
asked Nov 20, 2020 in Operating System jothee 27 views
0 votes
7 answers
36
NIELIT 2016 MAR Scientist C - Section C: 2
The file manager is responsible for naming files saving files deleting files all of these
The file manager is responsible for naming files saving files deleting files all of these
asked Apr 2, 2020 in Operating System Lakshman Patel RJIT 623 views
2 votes
4 answers
37
NIELIT 2016 MAR Scientist C - Section C: 7
In real-time operating systems, which of the following is the most suitable scheduling scheme? round-robin first-come-first-served preemptive random scheduling
In real-time operating systems, which of the following is the most suitable scheduling scheme? round-robin first-come-first-served preemptive random scheduling
asked Apr 2, 2020 in Operating System Lakshman Patel RJIT 485 views
2 votes
3 answers
38
NIELIT 2016 MAR Scientist C - Section C: 8
If there are $32$ segments, each of size $1$ K byte, then the logical address should have $13 \text{ bits}$ $14 \text{ bits}$ $15 \text{ bits}$ $16 \text{ bits}$
If there are $32$ segments, each of size $1$ K byte, then the logical address should have $13 \text{ bits}$ $14 \text{ bits}$ $15 \text{ bits}$ $16 \text{ bits}$
asked Apr 2, 2020 in Operating System Lakshman Patel RJIT 401 views
1 vote
2 answers
39
NIELIT 2016 MAR Scientist C - Section C: 26
The seek time of a disk is $30$ ms. It rotates at the rate of $30$ rotations/second. The capacity of each track is $300$ words. The access time is (approximately) $62$ ms $60$ ms $50$ ms $47$ ms
The seek time of a disk is $30$ ms. It rotates at the rate of $30$ rotations/second. The capacity of each track is $300$ words. The access time is (approximately) $62$ ms $60$ ms $50$ ms $47$ ms
asked Apr 2, 2020 in Operating System Lakshman Patel RJIT 238 views
2 votes
2 answers
40
NIELIT 2017 OCT Scientific Assistant A (IT) - Section C: 10
The address sequence generated by tracing a particular program executing in a pure demand paging system with $100$ records per page, with $1$ free main memory frame is recorded as follows. What is the number of Page Faults? $0100,0200,0430,0499,0510,0530,0560,0120,0220,0240,0260,0320,0370.$ $15,4$ $6,4$ $7,2$ $4,6$ [closed]
The address sequence generated by tracing a particular program executing in a pure demand paging system with $100$ records per page, with $1$ free main memory frame is recorded as follows. What is the number of Page Faults? $0100,0200,0430,0499,0510,0530,0560,0120,0220,0240,0260,0320,0370.$ $15,4$ $6,4$ $7,2$ $4,6$
asked Apr 1, 2020 in Operating System Lakshman Patel RJIT 224 views
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