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;
const int mod = 1e9+7;
int n,m;
char ss[200005];
int f[200005],qz[200005];
int main() {
	scanf("%d%d",&n,&m);
	scanf("%s",&ss[1]);
	bool mk = 0;
	int mx = 0;
	for(int i=1;i<=m;i++) {
		if(ss[i]!=ss[1]) { mk = 1; mx = i-1; break; }
	}
	if(!mk) {
		int ans = 1;
		f[0] = qz[0] = 1;
		for(int i=0;i<=n;i++) {
			if(i>=2) f[i] = qz[i-2];
			if(i!=0)qz[i] = (qz[i-1]+f[i])%mod;
			if(n-i>=2) ans = (ans+1ll*f[i]*(n-i)%mod)%mod;
		}
		printf("%d",ans);
		return 0;
	}
	if(n&1) {
		puts("0"); return 0;
	}
	int lx = 0;
	for(int i=1;i<=m;i++) {
		if(ss[i]==ss[1]) lx++;
		else {
			if(lx&1) mx = min(mx,lx);
			lx = 0;
		}
	}
	mx = mx/2+1;
	n/=2;
	f[0] = qz[0] = 1;
	int ans = 0;
	for(int i=0;i<=n;i++) {
		if(i) {
			f[i] = (qz[i-1] - (i-mx-1>=0?qz[i-mx-1]:0) )%mod;
			qz[i] = (qz[i-1]+f[i])%mod;
		}
		if(n-i<=mx) ans = (ans + 1ll * f[i]*2%mod*(n-i)%mod )%mod;
	}
	printf("%d",(ans%mod+mod)%mod );
}