Consider a circle whose perimeter is divided by N points into N arcs of equal length, and each of the arcs is painted red or blue. Such a circle is said to generate a string S from every point when the following condition is satisfied:

* We will arbitrarily choose one of the N points on the perimeter and place a piece on it.
* Then, we will perform the following move M times: move the piece clockwise or counter-clockwise to an adjacent point.
* Here, whatever point we choose initially, it is always possible to move the piece so that the color of the i-th arc the piece goes along is S_i, by properly deciding the directions of the moves.



Assume that, if S_i is `R`, it represents red; if S_i is `B`, it represents blue. Note that the directions of the moves can be decided separately for each choice of the initial point.

You are given a string S of length M consisting of `R` and `B`. Out of the 2^N ways to paint each of the arcs red or blue in a circle whose perimeter is divided into N arcs of equal length, find the number of ways resulting in a circle that generates S from every point, modulo 10^9+7.

Note that the rotations of the same coloring are also distinguished.

Constraints

* 2 \leq N \leq 2 \times 10^5
* 1 \leq M \leq 2 \times 10^5
* |S|=M
* S_i is `R` or `B`.

Input

Input is given from Standard Input in the following format:


N M
S


Output

Print the number of ways to paint each of the arcs that satisfy the condition, modulo 10^9+7.

Examples

Input

4 7
RBRRBRR


Output

2


Input

3 3
BBB


Output

4


Input

12 10
RRRRBRRRRB


Output

78
#include<cstdio>
#define N 200002
#define p 1000000007
inline int mod(int x){return x<p?x:x-p;}
int f[N],g[N],i,j,m,n,t,x;char s[N];
int main()
{
    scanf("%d%d%s",&n,&m,s+1),f[0]=g[0]=1;
    if(s[1]=='B')for(i=1;i<=m;i++)s[i]=s[i]=='B'?'R':'B';
    while(s[m]=='R')s[m--]=0;
    for(i=0;s[i+1]=='R';i++);
    for(t=i++|1;s[i];t=(i-j&1)&&i-j<t?i-j:t)
    {
        while(s[i]=='B')i++;
        for(j=i;s[i]=='R';i++);
    }
    if(m)for(i=1;i<=n;f[i]=mod(f[i-1]+g[i]),i++)if(~i&1)g[i]=i<t+2?f[i-1]:mod(p+f[i-1]-f[i-t-2]);else;
    else for(g[1]=1,i=2;i<=n;i++)g[i]=mod(g[i-2]+g[i-1]);
    if(m)for(i=n-2;0<=i&&n-i<=t+1;i-=2)x=(x+(long long)g[i]*(n-i))%p;
    else x=mod(g[n]+g[n-2]);
    return 0*printf("%d\n",x);
}