There is a perfect binary tree with $2^{10^{100}}-1$ vertices, numbered $1,2,...,2^{10^{100}}-1$.  
Vertex $1$ is the root. For each $1\leq i < 2^{10^{100}-1}$, Vertex $i$ has two children: Vertex $2i$ to the left and Vertex $2i+1$ to the right.

Takahashi starts at Vertex $X$ and performs $N$ moves, represented by a string $S$. The $i$-th move is as follows.

-   If the $i$-th character of $S$ is `U`, go to the parent of the vertex he is on now.
-   If the $i$-th character of $S$ is `L`, go to the left child of the vertex he is on now.
-   If the $i$-th character of $S$ is `R`, go to the right child of the vertex he is on now.

Find the index of the vertex Takahashi will be on after $N$ moves. In the given cases, it is guaranteed that the answer is at most $10^{18}$.

Input is given from Standard Input in the following format:

```
$N$ $X$
$S$
```

-   $1 \leq N \leq 10^6$
-   $1 \leq X \leq 10^{18}$
-   $N$ and $X$ are integers.
-   $S$ has a length of $N$ and consists of `U`, `L`, and `R`.
-   When Takahashi is at the root, he never attempts to go to the parent.
-   When Takahashi is at a leaf, he never attempts to go to a child.
-   The index of the vertex Takahashi is on after $N$ moves is at most $10^{18}$.

6

The perfect binary tree has the following structure.

In the three moves, Takahashi goes 2→1→3→6.

500000000000000000

During the process, Takahashi may be at a vertex whose index exceeds $10^{18}$.