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 N = 2e5+5;
const int mod = 1e9+7;
inline int add(int a,int b){a+=b;return a>=mod?a-mod:a;}
char s[N];
int n,m,mx;
int f[N],f2[N],ans;

int main()
{
	cin >> n >> m;
	scanf("%s",s+1);
	int p;
	for(p=1;p<=m;p++){
		if(s[p]!=s[1])break;
	}
	mx=p-1;
	if(mx==m){
		f[0] = f2[0] = ans = 1;
		for(int i=0;i<=n;i++){
			if(i>=2)f[i] = f2[i-2];
			f2[i] = add((i-1>=0?f2[i-1]:0),f[i]);
			if(n-i>=2)ans = add(ans, 1ll*f[i]*(n-i)%mod);
		}
		cout << ans << endl;
		return 0;
	}
	if(n&1){
		puts("0");
		return 0;
	}
	if(mx%2==0)++mx;
	for(int t=0;p<=m;p++){
		if(s[p]==s[1])++t;
		else{
			if(t&1)mx = min(mx, t);
			t=0;
		}
	}
	n/=2,mx=(mx+1)/2;
	f[0]=f2[0]=1;
	for(int i=0;i<=n;i++)
	{
		if(i)
		{
			f[i]=(mod+f2[i-1]-(i-mx-1>=0?f2[i-mx-1]:0))%mod;
			f2[i]=(f2[i-1]+f[i])%mod;
		}
		if(n-i<=mx) ans = add(ans,(mod+1ll*(n-i)*f[i]*2%mod)%mod);
	}
	cout<<ans;
}