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 llong long long
using namespace std;

inline int read()
{
	int x = 0,f = 1; char ch = getchar();
	for(;!isdigit(ch);ch=getchar()) {if(ch=='-') f = -1;}
	for(; isdigit(ch);ch=getchar()) {x = x*10+ch-48;}
	return x*f;
}

const int N = 2e5;
const int P = 1e9+7;
char a[N+3];
llong f[N+3],sf[N+3];
int n,m; llong ans;

int main()
{
	scanf("%d%d",&n,&m); scanf("%s",a+1);
	for(int i=1; i<=m; i++) a[i] = a[i]=='R'?0:1;
	if(a[1]) {for(int i=1; i<=m; i++) a[i] ^= 1;}
	int lim = m,cur = 0;
	for(int i=1; i<=m; i++)
	{
		if(a[i]==a[1]) {cur++;}
		else
		{
			if(cur==i-1) {lim = cur&1?cur:cur+1;}
			else {if(cur&1) lim = min(lim,cur);}
			cur = 0;
		}
	}
	if(cur==m)
	{
		f[1] = 1ll; for(int i=2; i<=n; i++) f[i] = (f[i-1]+f[i-2])%P;
		ans = (f[n]+f[n-1]+f[n-1])%P;
		printf("%lld\n",ans);
	}
	else
	{
		if(n&1) {puts("0"); return 0;}
//		printf("lim=%d\n",lim);
		lim>>=1,n>>=1; f[1] = sf[1] = 1ll;
		for(int i=2; i<=n; i++)
		{
			f[i] = (sf[i-1]-sf[max(0,i-lim-2)]+P)%P; sf[i] = (sf[i-1]+f[i])%P;
		}
		for(int i=1; i<=n; i++) if(n-i<=lim)
		{
			ans = (ans+f[i]*(n-i+1ll))%P;
		}
		printf("%lld\n",(ans+ans)%P);
	}
	return 0;
}