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
using namespace std;
const int N = 501000,mod = 1e9+7;
char s[N];
int a[N],n,m;
void solve1(){
	int f[N][2][2];memset(f,0,sizeof(f));
	f[1][1][1]=f[1][0][0]=1;
	auto upd=[](int&a,int b){a+=b;if(a>=mod)a-=mod;};
	for(int i=2;i<=n;i++){
		upd(f[i][1][1],f[i-1][1][0]);
		upd(f[i][1][1],f[i-1][1][1]);
		upd(f[i][1][0],f[i-1][1][1]);
		upd(f[i][0][1],f[i-1][0][0]);
		upd(f[i][0][1],f[i-1][0][1]);
		upd(f[i][0][0],f[i-1][0][1]);
	}
	cout<<(0ll+f[n][1][1]+f[n][1][0]+f[n][0][1])%mod<<'\n';
}
void solve2(){
	int _f[N],*f;memset(_f,0,sizeof(_f));
	f=_f+100;
	int len=n|1,now=0,pos=m;
	while(a[pos]) pos--;
	for(int i=pos;i;i--)
		if(!a[i]){
			if(now&1) len=min(len,now);
			now=0;
		}
		else now++;
	len=min(len,now|1);
	f[-1]=f[1]=1;
	for(int i=3;i<=n;i+=2){
		f[i]=2ll*f[i-2]%mod;
		if(i>len) f[i]=(f[i]-f[i-len-3]+mod)%mod;
	}
	ll ans=0;
	for(int i=1;i<=len;i+=2) ans=(1ll*(i+1)*f[n-i-2]+ans)%mod;
	cout<<ans<<'\n';
}
int main(){
	ios::sync_with_stdio(0);
	cin>>n>>m>>s+1;
	for(int i=1;i<=m;i++)
		a[i]=s[i]==s[1];
	int flag=1;
	for(int i=1;i<=m;i++)
		flag&=a[i];
	if(flag) solve1();
	else solve2();
	return 0;
}