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<map>
#include<set>
#include<cmath>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 200010
#define int long long
#define mod 1000000007
#define debug cerr<<__LINE__<<" "<<__FUNCTION__<<"\n"

inline int read(){
    int x=0,y=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')y=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    return x*y;
}
void put(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>=10) put(x/10);
    putchar((x%10)+48);
}
int n,m,A[N],B[N];
string s;
signed main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	n=read();m=read();
	cin>>s;
	if(s[0]=='B'){
		for(int i=0;i<m;i++) s[i]=(s[i]=='B'?'R':'B');
	}
	int pos=s.find('B');
	if(pos==string::npos){
		A[1]=A[2]=1;
		for(int i=3;i<=n;i++) A[i]=(A[i-1]+A[i-2])%mod;
		cout<<(A[n]+2*A[n-1]%mod)%mod<<"\n";
		return 0;
	}
	if(n&1){
		puts("0");
		return 0;
	}
	int l=0x7ffffffff,res=0,flag=1;
	for(int i=0;i<m;i++){
		if(s[i]=='B'){
			if(flag||(res&1)) l=min(l,res+2);
			flag=0;
			res=0;
		}else res++;
	}
//	cerr<<l<<"\n";
	l/=2;n/=2;
	A[0]=B[0]=1;
	for(int i=1;i<=n;i++){
		A[i]=B[i-1];
		if(i>l) A[i]=(A[i]-B[i-l-1]+mod)%mod;
		B[i]=(B[i-1]+A[i])%mod;
	}
	int ans=0;
	for(int i=1;i<=l;i++){
		ans=(ans+2*i%mod*A[n-i]%mod)%mod;
	}
	cout<<ans<<"\n";
//	fclose(stdin);
//	fclose(stdout);
	return 0;
}
