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 mod 1000000007
#define MN 200000
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
char s[MN+5];
int g[MN+5],n,m,R,all=1;
void Dp1()//每个都有一个合法相连
{
    g[1]=1;int ans=1;
    for(int i=2;i<=n;++i) g[i]=(g[i-1]+g[i-2])%mod;
    for(int i=1;i<n;++i) ans=(ans+1LL*(g[i]-g[i-1]+mod)*(n-i+1))%mod;
    printf("%d",ans);
}
void Dp2()//长度为奇数且处于[1,R]
{
    int ans=0,first=1,R=n-1;
    for(int i=1,j;i<=m;i=j)
    {
        for(j=i;j<=m&&s[j]==s[1];++j);
        if(j>m) break;
        int len=j-i;
        if(len&1) R=min(R,len);
        else first?R=min(R,len+1):0;
        first=0;
        for(;j<=m&&s[j]!=s[1];++j);
    }
    g[1]=1;
    for(int i=3;i<=n;i+=2)
    {
        g[i]=(g[i-2]-g[max(0,i-R-3)]+mod)%mod;
        g[i]=(g[i]+g[i-2])%mod;
    }
    for(int i=max(1,n-R);i<=n;i+=2) ans=(ans+1LL*(n-i+1)*(g[i]-g[max(0,i-2)]+mod))%mod;
    printf("%d",ans);
}
int main()
{
    R=n=read();m=read();scanf("%s",s+1);
    for(int i=1;i<=m;++i) if(s[i]!=s[1]) all=0;
    if(all) Dp1();
    else (n&1)?puts("0"):(Dp2(),0);
    return 0;
}