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<bits/stdc++.h>
const int mo=1e9+7,N=2e5+5;
int n,m,i,j,ans,lm=N,f[N],s[N];
char c[N];
int main(){
	scanf("%d%d%s",&n,&m,c+1);
	for(i=1;c[i]==c[1];++i);
	if(i>m){
		static int f[N][2];
		f[0][0]=1;
		for(i=1;i<=n;++i)f[i][0]=(f[i-1][0]+f[i-1][1])%mo,f[i][1]=f[i-1][0];
		ans=f[n][0];
		memset(f,0,sizeof f);
		f[0][1]=1;
		for(i=1;i<=n;++i)f[i][0]=(f[i-1][0]+f[i-1][1])%mo,f[i][1]=f[i-1][0];
		ans=(ans+f[n][1])%mo;
	}else{
		lm=i&1?i:i-1;
		for(;i<=m;i=j){
			for(j=i+1;c[j]==c[1];++j);
			if(j<=m && (j-i-1)%2==1)lm=std::min(lm,j-i-1);
		}
		if(n&1){puts("0");return 0;}
		lm-=lm%2==0;lm=lm+1>>1;
		s[0]=f[0]=1;
		for(i=1;i<=n;++i)f[i]=(s[i-1]+mo-(i>lm?s[i-lm-1]:0))%mo,s[i]=(s[i-1]+f[i])%mo;
		for(i=1;i<=lm*2-1 && i+1<=n;i+=2)ans=(ans+1ll*(i+1)*f[n-i-1>>1])%mo;
	}
	printf("%d\n",ans);
	return 0;
}