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<iostream>
#include<cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=200200,P=1e9+7;
int n,m,ans,lim,all,mxo;
int b[N],f[N],s[N];
char S[N];
void work(int w){
	f[0]=s[0]=1;//cout<<w<<' '<<lim<<'\n';
	if(w==1){
		f[1]=0;s[1]=1;
		FOR(i,2,n){
			f[i]=s[i-2];
			s[i]=(s[i-1]+f[i])%P;
			//cout<<f[i]<<' ';
		}
	}
	else{
		FOR(i,1,n){
			f[i]=(s[i-1]-(i-lim-1<0?0:s[i-lim-1])+P)%P;
			s[i]=(s[i-1]+f[i])%P;
			//cout<<f[i]<<'\n';
		}
	}
	FOR(i,1+(w==1),min(lim,n)) (ans+=1ll*i*f[n-i]%P)%=P;
	cout<<w*ans%P<<'\n';
}
int main(){
	//freopen("1.in","r",stdin);
	scanf("%d%d%s",&n,&m,S+1);
	ROF(i,m,1) b[i]=S[i]=='R'^S[1]=='B';
	all=1;
	FOR(i,1,m) all&=b[i];
	if(all) return lim=n,ans=1,work(1),0;
	FOR(i,1,m)if(!b[i]){lim=i-!(i%2);break;}
	for(int l=1,r=1;l<=m;l=r){
		while(r<=m && b[r]==b[l]) r++;
		if(b[l]==0) continue;
		if((r-l&1) && l!=1 && r<=m) lim=min(lim,r-l);
	}
	lim=lim+1>>1;
	if(n&1) return puts("0"),0;
	n/=2;work(2);
}
