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>
#include<algorithm>
using namespace std;
const int N=2e5+5;
const int mod=1e9+7;
int n,m,k,f[N],ans;char str[N];
int main(){
	scanf("%d%d%s",&n,&m,str+1);
	k=n;while(m&&str[m]==str[1])--m;
	if(!m)ans=1;
	for(int i=1,j=1;i<=m;i=j=j+1)
		if(str[i]==str[1]){
			while(j<m&&str[j+1]==str[1])++j;
			if(i==1)
				if(j-i+1&1)k=min(k,j-i+1);
				else k=min(k,j-i+2);
			else
				if(j-i+1&1)k=min(k,j-i+1);
		}
	if(k==n){
		for(int i=f[0]=1,j=0;i<=n;++i){
			f[i]=j;j=(j+f[i-1])%mod;
		}
		for(int i=2;i<=n;++i)
			ans=(ans+1ll*i*f[n-i])%mod;
	}else{
		++k;
		for(int i=2,j=f[0]=1;i<=n;i+=2){
			f[i]=j;j=(j+f[i])%mod;
			if(i>=k)j=(j-f[i-k]+mod)%mod;
		}
		for(int i=2;i<=k;i+=2)
			ans=(ans+1ll*i*f[n-i])%mod;
	}
	printf("%d\n",ans);
	return 0;
}
