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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod=1e9+7;
char s[200005];
int f[200005];
int main(){
	// freopen("b.in","r",stdin);
	int n,m;
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	if(s[1]=='B')
		for(int i=1;i<=m;i++)s[i]=(s[i]=='R'?'B':'R');
	int lst=0,k=n&1?n-2:n-1;
	for(int i=1;i<=m;i++){
		if(s[i]=='B'){
			if(lst!=i-1){
				if(!lst){
					if((i-lst-1)&1)k=min(k,i-lst-1);
					else k=min(k,i-lst-1+1);
				}
				if((i-lst-1)&1)k=min(k,i-lst-1);
			}
			lst=i;
		}
	}
	if(!lst){
		int ans=2;
		f[1]=1;
		for(int i=2;i<=n;i++){
			f[i]=(f[i-1]+f[i-2])%mod;
			if(i!=n)ans=(ans+f[i])%mod;
			else ans=(ans+f[i-1])%mod;
		}
		// puts("ok");
		printf("%d\n",ans);
		return 0;
	}
	k++;
	f[1]=f[3]=1;
	for(int i=5;i<=n;i+=2)f[i]=(2ll*f[i-2]+(i>=k+2?mod-f[i-k-2]:0))%mod;	
	int ans=0;
	for(int i=1;i<=min(n,k);i++)if(i&1)ans=(ans+1ll*f[n-i]*(i+1))%mod;
	printf("%d\n",ans);
	return 0;
}