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<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const ll N=2e5+10,mo=1e9+7;
ll f[N],n,m,lim,q[N];
char s[N];
void move(ll &a,ll b){a=(a+b)%mo;}
int main(){
	cin>>n>>m;
	scanf("%s",s+1);
	if (s[1]=='B'){
		for (ll i=1;i<=m;i++)s[i]='B'+'R'-s[i];
	}
	ll cnt=0;
	for (ll i=1;i<=m;i++){
		if (s[i]=='R')cnt++;
		else {
			if (!lim)lim=cnt|1;
			else if (cnt&1)lim=min(lim,cnt);
			cnt=0;
		}
	}
	if (!lim){
		for (int i=1;i<=n;i++){
			move(f[i],q[i-2]+i-(i==1));
			q[i]=(q[i-1]+f[i])%mo;
		}
		cout<<f[n]+1<<endl;
		return 0;
	}
	lim++;
	for (ll i=1;i<=n;i++){
		f[i]=q[i-2];
		if (i<=lim&&(i%2==0))move(f[i],i);
		q[i]=((ll)q[i-2]+f[i]-(i>lim?f[i-lim]:0)+mo)%mo;
	}
	cout<<f[n]<<endl;
}