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<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int Mod=1000000007;
int a[200010];
char s[200010];
int f[200010],g[200010];
int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	scanf("%s",s+1);
	for(int i=1;i<=m;i++)
		a[i]=s[i]=='B';
	if(a[1]){
		for(int i=1;i<=m;i++)
			a[i]^=1;
	}
	bool flag=true;
	for(int i=1;i<=m;i++)
		if(a[i])flag=false;
	if(flag){
		f[0]=1;
		g[0]=1;
		for(int i=1;i<=n;i++)
		{
			if(i>1)f[i]=g[i-2];
			g[i]=(g[i-1]+f[i])%Mod;
		}
		int ans=1;
		for(int i=2;i<=n;i++)
			ans=(ans+1ll*f[n-i]*i)%Mod;
		printf("%d\n",ans);
		return 0;
	}
	if(n&1){
		printf("0\n");
		return 0;
	}
	int lim=n;
	int lst=0;
	for(int i=1;i<=m;i++)
		if(a[i]){
			if(!lst)lim=min(lim,i-lst+1);
			else if((i-lst)%2==0)lim=min(lim,i-lst);
			lst=i;
		}
	lim/=2;
	f[0]=1;
	g[0]=1;
	for(int i=1;i<=n/2;i++)
	{
		f[i]=g[i-1];
		if(i-lim>0)f[i]=(f[i]-g[i-lim-1]+Mod)%Mod;
		g[i]=(g[i-1]+f[i])%Mod;
	}
	int ans=0;
	for(int i=1;i<=lim;i++)
		ans=(ans+1ll*i*2*f[n/2-i])%Mod;
	printf("%d\n",ans);
	return 0;
}