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=200100,mod=1000000007;
typedef long long ll;
int f[N],sum[N],n,m;
char s[N];

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m>>(s+1);
	int mn=max(n,m)+1,sz=0,flag=0;
	for(int i=1;i<=m;++i)	
		if(s[i]==s[1])++sz;
		else{
			if(sz&1)mn=min(mn,sz);
			else if(sz&&!flag)mn=min(mn,sz+1);
			sz=0;flag=1;
		}
	int L=0,R=0;
	if(flag){
		if(n&1){
			cout<<0<<'\n';
			return 0;
		}
		n/=2;L=1;R=min(n,(mn+1)/2);
	}else L=2,R=n;
	f[0]=sum[0]=1;
	for(int i=1;i<=n;++i){
		f[i]=sum[i-L];if(i>R)f[i]=(f[i]-sum[i-R-1]+mod)%mod;
		sum[i]=(sum[i-1]+f[i])%mod;
	}
	int ans=0;
	for(int i=L;i<=R;++i)ans=(ans+(ll)i*f[n-i])%mod;
	if(flag)ans=ans*2%mod;else ans=(ans+1)%mod;
	cout<<ans<<'\n';
	return 0;
}
