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;
#define ll long long

const int mod=1e9+7;

void U(int &x,int y){
    x=(x+y)%mod;
}

int n, m, a[202000];
char s[202000];

int main(){
    cin>>n>>m>>s+1;
    for (int i=1;i<=m;++i)
        a[i]= s[i]==s[1];
    if (*min_element(a+1,a+m+1)==1){
        static int f[202000][2][2];
        f[1][0][0]=f[1][1][1]=1;
        for (int i=1;i<n;++i){
            for (int s=0;s<=1;++s)
                for (int t=0;t<=1;++t)
                    for (int tt=0;tt<=1;++tt){
                        if (t==0&&tt==0) continue;
                        U(f[i+1][s][tt],f[i][s][t]);
                    }
        }
        int ans=(1ll*f[n][0][1]+f[n][1][0]+f[n][1][1])%mod;
        cout<<ans<<endl;
        return 0;
    }
    int las=0, mx=1e9; vector<int>vec;
    for (int i=1;i<=m;++i)
        if (a[i]){
            ++las;
        }else{
            vec.push_back(las);
            las=0;
        }
    {
        int t=vec.front(); vec.erase(vec.begin());
        mx=min(mx,(t&1? t: t+1));
    }
    for (auto t:vec){
        if (t&1) mx=min(mx,t);
    }


    static int ans=0, f[202000], sum[202000];
    f[1]=sum[1]=1;
    for (int i=1;i<=n;++i){
        if (i!=1){
            f[i]=(sum[i-2]-sum[max(0,i-mx-3)]+mod)%mod;
            sum[i]=(sum[i-2]+f[i])%mod;
        }
        if ((n-i&1)&&n-i<=mx){
            ans=(ans+(ll)f[i]*(n-i+1))%mod;
        }
    }
    cout<<ans<<endl;
}
