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 fi first
#define se second
#define pb push_back
#define SZ(x) ((int)x.size())
#define L(i,u) for (register int i=head[u]; i; i=nxt[i])
#define rep(i,a,b) for (register int i=(a); i<=(b); i++)
#define per(i,a,b) for (register int i=(a); i>=(b); i--)
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef pair<int,int> Pii;
typedef vector<int> Vi;
template<class T> inline void read(T &x){
	x=0; char c=getchar(); int f=1;
	while (!isdigit(c)) {if (c=='-') f=-1; c=getchar();}
	while (isdigit(c)) {x=x*10+c-'0'; c=getchar();} x*=f;
}
template<class T> inline void umin(T &x, T y){x=x<y?x:y;}
template<class T> inline void umax(T &x, T y){x=x>y?x:y;}
inline ui R() {
	static ui seed=416;
	return seed^=seed>>5,seed^=seed<<17,seed^=seed>>13;
}
const int N = 233333,mo=1e9+7;
int n,m,f[N],qz[N];char s[N];
void init(int n, int lim){//<=lim
	umin(lim,n);//f[i]=sigma f[i-1..i-lim]
	f[0]=1;qz[0]=1;
	rep(i,1,n){
		f[i]=(qz[i-1]-(i-lim-1>=0?qz[i-lim-1]:0)+mo)%mo;
		qz[i]=(qz[i-1]+f[i])%mo;
	}
}
int main() {
	read(n);read(m);scanf("%s",s+1);
	int cnt=0;rep(i,1,m)cnt+=s[i]=='R';
	if(cnt==0||cnt==m){
		static int f[N];
		f[0]=1;f[1]=2;rep(i,2,n)f[i]=(f[i-1]+f[i-2])%mo;
		printf("%d\n",(f[n-1]+(n-3>=0?f[n-3]:1))%mo);
		return 0;
	}
	if(n&1){puts("0");return 0;}
	int lim=1e9;int j;for(j=1;j<=m;j++)if(s[j]!=s[1])break;j--;
	if(j%2==0)lim=j+1;
	rep(i,1,m)if(s[i]==s[1]){
		int j;for(j=i;j<=m;j++)if(s[j]!=s[i])break;
		//[i,j-1]
		if((j-i)%2&&j-1<m)umin(lim,j-i);
		i=j-1;
	}
//	cerr<<lim<<endl;
	lim++;init(n/2,lim/2);int res=0;//%2==0
	for(int i=2;i<=min(lim,n);i+=2)res=(res+1LL*i*f[(n-i)/2])%mo;
	cout<<res;
	return 0;
}
