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 rep(i,n) for ((i)=1;(i)<=(n);(i)++)
using namespace std;
const int mod=1e9+7;
int n,m,i,lst,dp[200005],s[200005],f[200005][2],tmp,mi=0x3f3f3f3f,l;
string st;
int main(){
	cin>>n>>m;
	cin>>st;
	st=" "+st;
	i=1;lst=1;
	while(i<=m){
		if(st[i]!=st[i-1]){
			if(st[i]!=st[1]){
				if(lst==1||((i-lst)&1)){
					mi=min(mi,i-lst);
				}
			}
			else{
				lst=i;
			}
		}
		i++;
	}
	if(mi<0x3f3f3f3f){
		if(n&1){
			puts("0");
			return 0;
		}
		mi=mi/2+1;
		n/=2;
		l=1;
/*		s[1]=1;s[mi+1]+=mod-1;
		rep(i,n){
			tmp=(tmp+s[i])%mod;
			dp[i]=tmp;
			s[min(n,i+1)]=(s[min(n,i+1)]+dp[i])%mod;
			s[min(n,i+mi)+1]=(s[min(n,i+mi)+1]+mod-dp[i])%mod;
		}
		cout<<dp[n]<<endl;*/
	}
	else{
		l=2;
	}
	for(i=l;i<=n;i++){
		dp[i]=(s[i-l]+(i>mi?-s[i-mi-1]:(3ll-l)*i))%mod;
		s[i]=(s[i-1]+dp[i])%mod;
	}
	cout<<(dp[n]+l-1+mod)%mod<<endl;
	return 0;
}