A single tape Turing Machine $M$ has two states $q0$ and $q1$, of which $q0$ is the starting state. The tape alphabet of $M$ is $\{0, 1, B\}$ and its input alphabet is $\{0, 1\}$. The symbol $B$ is the blank symbol used to indicate end of an input string. The transition function of $M$ is described in the following table.

$$\begin{array}{|l|l|l|l|}\hline \text{} & \text{$0$} & \text{$1$} & \text{$B$} \\\hline \text{$q0$} & \text{$q1, 1, R$} & \text{$q1, 1, R$} & \text{$Halt$} \\\hline \text{$q1$} & \text{$q1, 1, R$} & \text{$q0, 1, L$} & \text{$q0, B, L$} \\\hline \end{array}$$

The table is interpreted as illustrated below.

The entry $(q1, 1, R)$ in row $q0$ and column $1$ signifies that if $M$ is in state $q0$ and reads $1$ on the current page square, then it writes $1$ on the same tape square, moves its tape head one position to the right and transitions to state $q1$.

Which of the following statements is true about $M$?

- $M$ does not halt on any string in $(0+1)^+$
- $M$ does not halt on any string in $(00+1)^*$
- $M$ halts on all strings ending in a $0$
- $M$ halts on all strings ending in a $1$