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>
#define ll long long 
#define mod 1000000007

using namespace std;
const int N=200010;
int n,m,ans,f[N],g[N];
char s[N];

void inc(int&x,int y){x+=y;if(x>=mod)x-=mod;}
void dec(int&x,int y){x-=y;if(x<0)x+=mod;}

int main(){
//	freopen("E.in","r",stdin);
//	freopen("E.out","w",stdout);
	scanf("%d%d%s",&n,&m,s+1);
	int p=0;while(p<m&&s[p+1]==s[1])++p;
	if(p==m){
		ans=f[0]=g[0]=1;
		for(int i=0;i<=n;++i){
			if(i>1)f[i]=g[i-2];
			if(i)g[i]=(g[i-1]+f[i])%mod;
			if(n-i!=1)inc(ans,(ll)(n-i)*f[i]%mod);
		}
		cout<<ans<<endl;
	}else{
		if(n&1)puts("0"),exit(0);
		int L=p|1;
		while(1){
			int tmp=0;
			while(p<m&&s[p+1]!=s[1])++p;tmp-=p;
			while(p<m&&s[p+1]==s[1])++p;tmp+=p;
			if(p==m)break;
			if(tmp&1)L=min(L,tmp);
		}
		f[0]=g[0]=1;
		L=(L+1)>>1;n>>=1;
		for(int i=0;i<=n;++i){
			if(i)f[i]=g[i-1];
			if(i>L)dec(f[i],g[i-L-1]);
			if(i)g[i]=(g[i-1]+f[i])%mod;
			if(n-i<=L)inc(ans,(ll)(n-i)*f[i]%mod);
		}
		cout<<ans*2%mod<<endl;
	}
	return 0;
} 
