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>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

typedef long long ll;

const int N=2e5+5,Mo=1e9+7;

int n,m,k,f[N],s[N];
char st[N];

int main() {
	scanf("%d%d",&n,&m);
	scanf("%s",st+1);
	k=1;for(;k<m&&st[k+1]==st[1];k++);
	if (k==m) {
		f[0]=s[0]=1;
		fo(i,1,n) {
			if (i>=2) f[i]=s[i-2];
			s[i]=(s[i-1]+f[i])%Mo;
		}
		int ans=1;
		fo(i,2,n) (ans+=(ll)i*f[n-i]%Mo)%=Mo;
		printf("%d\n",ans);
		return 0;
	}
	if (n&1) {puts("0");return 0;}
	int lim=k+1,now=0;
	for(k++;k<=m;k++) {
		if (st[k]==st[1]) now++;
		else {
			if (now&1) lim=min(lim,now);
			now=0;
		}
	}
	n>>=1;lim=(lim+1)>>1;
	f[0]=s[0]=1;
	fo(i,1,n) {
		f[i]=s[i-1];
		if (i-lim-1>=0) (f[i]+=Mo-s[i-lim-1])%=Mo;
		s[i]=(s[i-1]+f[i])%Mo;
	}
	int ans=0;
	fo(i,1,lim) (ans+=(ll)(i<<1)*f[n-i]%Mo)%=Mo;
	printf("%d\n",ans);
	return 0;
}