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;

typedef long long LL;
#define N 300000

const LL mod=1000000007;
LL n,m,p,ans,f[N];
char s[N];

void upd(LL &x,LL y){x=(x+y)%mod;}

int main(){
	scanf("%lld%lld%s",&n,&m,s+1);
	if (s[1]=='B') for (LL i=1;i<=m;++i) s[i]=s[i]=='B'?'R':'B';
	p=n;
	ans=1;
	for (LL i=1,j=0,fst=0;i<=m;++i){
		if (s[i]=='R') ++j;
		else{
			ans=0;
			if (!fst) p=min(p,j&1?j:j+1);
			fst=1;
			if (j&1) p=min(p,j);
			j=0;
		}
	}
	if (!ans&&(n&1)){puts("0"); return 0;}
	if (ans){
		ans=0;
		f[1]=1;
		for (LL i=2;i<=n;++i) f[i]=(f[i-1]+f[i-2])%mod;
		upd(ans,f[n]+f[n-1]*2);
	}
	else{
		p=(p+1)/2;
		f[0]=1;
		for (LL i=1,j=0,k=0;i<=n/2;++i){
			upd(k,f[i-1]);
			if (j<i-p){upd(k,-f[j]); ++j;}
			upd(f[i],k);
		}
		for (LL i=2;i<=n&&i<=p*2;i+=2) upd(ans,i*f[(n-i)/2]);
	}
	printf("%lld\n",(ans+mod)%mod);
	
	return 0;
}
			