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 ll long long
#define mo 1000000007
#define N 200010
using namespace std;
int n,m,a[N],b[N],nb;
ll dp[N],s[N],ans;
int main(){
	cin>>n>>m;
	for (int i=1;i<=m;i++){char ch=getchar();while (ch!='R'&&ch!='B')ch=getchar();a[i]=(ch=='R');}
	for (int la=0,i=1;i<=m;i++)if (a[i]!=a[1]){b[++nb]=i-la-1;la=i;}
	if (nb==0){
		s[0]=s[1]=ans=1;
		for (int i=2;i<n;i++)s[i]=(s[i-1]+s[i-2])%mo;
		for (int j=0;j<n-1;j++)ans=(ans+s[j])%mo;ans=(ans+s[n-2])%mo;
	}else{
		int ma=b[1];
		for (int i=2;i<=nb;i++)if (b[i]%2)ma=min(ma,b[i]);
		if (ma%2==0)ma++;ma++;
		dp[0]=1;s[0]=1;
		for (int i=1;i<=n;i++){
			if (i%2)dp[i]=0;
			else{dp[i]=s[i-1];if (i-ma-1>=0)dp[i]=(dp[i]-s[i-ma-1]+mo)%mo;}
			s[i]=(s[i-1]+dp[i])%mo;
		}
		for (int j=n-2;j>=0;j-=2)if (n-j<=ma)ans=(ans+dp[j]*(n-j))%mo;
	}
	cout<<ans<<endl;
	return 0;
}