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>
using namespace std;
#define int long long
const int N=200005,M=1e9+7;
int n,m,f[N],g[N],ans;
char s[N];
signed main(){
	scanf("%lld%lld",&n,&m);
	scanf("%s",s+1);
	if (s[1]=='R'){
		for (int i=1;i<=m;i++){
			if (s[i]=='R')s[i]='B';
			else s[i]='R';
		}
	}
	int l=0,mi=0;
	for (int i=1;i<=m;i++)
		if (s[i]=='R'){
			l=i-1;
			break;
		}
	if (!l){
		f[0]=g[0]=1;
		for (int i=1;i<=n;i++){
			f[i]=g[i-2];
			g[i]=(g[i-1]+f[i])%M;
		}
		for (int i=2;i<=n;i++)(ans+=i*f[n-i])%=M;
		printf("%lld\n",(ans+1)%M);
		return 0;
	}
	mi=l+!(l%2);
	for (int i=1,j=1;i<=m;i=j){
		while (j<=m&&s[j]=='B')j++;
		if (j>m)break;
		if ((j-i)&1)mi=min(mi,j-i);
		while (j<=m&&s[j]=='R')j++;
	}
	if (n%2==1){
		puts("0");
		return 0;
	}
	mi=(mi+1)/2;
	n/=2;
	f[0]=g[0]=1;
	for (int i=1;i<=n;i++){
		f[i]=(g[i-1]-(i-mi-1<0?0:g[i-mi-1])+M)%M;
		g[i]=(g[i-1]+f[i])%M;
	}
	for (int i=1;i<=mi;i++)(ans+=i*2*f[n-i])%=M;
	printf("%lld\n",ans);
}