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 Fail puts("NO"),exit(0);
#define eps 1e-10
#define maxn 200010
#define maxm 100010
#define inf 1000000007
#define mod 1000000007
#define pi acos(-1)
#define mp(x,y) make_pair(x,y)
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int>pii;
typedef bitset<maxn> bit;
int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch-'0'<0||ch-'0'>9){if(ch=='-') f=-1;ch=getchar();}
    while(ch-'0'>=0&&ch-'0'<=9){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m;
char s[maxn];
int a[maxn];
bool check()
{
    for(int i=1;i<=m;i++)  if(a[i])  return false;
    return true;
}
int g[maxn],f[maxn],sum[maxn][2];
int L=inf;
int main()
{
    n=read();m=read();
    scanf("%s",s+1);
    for(int i=1;i<=m;i++)  a[i]=(s[i]=='B');
    if(a[1])  for(int i=1;i<=m;i++)  a[i]^=1;
    if(check())
    {
        if(n==2) return puts("3"),0;
        g[0]=1;g[1]=2;
        for(int i=2;i<=n;i++)  g[i]=(g[i-1]+g[i-2])%mod;
        return printf("%d\n",(g[n-1]+g[n-3])%mod),0;
    }
    for(int i=1;i<=m;i++)
    {
        if(a[i])  continue;
        int R=i;
        while(R<m&&!a[R+1])  R++;
        if(R==m)  break;
        if(i==1)  L=min(L,R-i+2);
        if((R-i+1)&1)  L=min(L,R-i+1);
        i=R;
    }
    f[0]=1;sum[0][0]=1;
    for(int i=1;i<=n;i++)
    {
        if(i>=2)
        {
            f[i]=sum[i-2][i&1];
            if(i>=L+2)  f[i]=(f[i]-sum[i-L-2][i&1]+mod)%mod;
        }
        sum[i][0]=sum[i-1][0];sum[i][1]=sum[i-1][1];
        sum[i][i&1]=(sum[i][i&1]+f[i])%mod;
    }
    int ans=f[n];
    for(int i=1;i<=L;i+=2)  ans=(ans+1ll*i*f[n-i-1])%mod;
    cout<<ans<<endl;
    return 0;
}