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

using namespace std;

typedef long long ll;

int f[200005];

int solve(int n,int up) {
  int s=0;
  for(int i=1;i<=n;i++) {
  	if (i>up+1) s=(s-f[i-up-1]+MOD)%MOD;
  	f[i]=(s+((i<=up)?i:0))%MOD;
  	s=(s+f[i])%MOD;
  }
  return f[n];
}

int solve2(int n) {
  f[1]=1;
  for(int i=2;i<=n+1;i++) f[i]=(f[i-1]+f[i-2])%MOD;
  return (f[n+1]+f[n-1])%MOD;
}

char str[200005];

int main() {
  int n,m;
  scanf("%d%d%s",&n,&m,str+1);
  if (str[1]=='B') {
  	for(int i=1;i<=m;i++) str[i]=((str[i]=='R')?'B':'R');
  }
  bool v=0;
  for(int i=1;i<=m;i++)
    if (str[i]=='B') {
    	v=1;
    	break;
	}
  if (!v) {
  	printf("%d\n",solve2(n));
  	return 0;
  }
  if (n&1) {
  	puts("0");
  	return 0;
  }
  int minn=n,s=0;
  bool fir=1;
  for(int i=1;i<=m;i++)
    if (str[i]=='R') s++;
    else {
    	if (s&&(fir||(s&1))) minn=min(minn,s);
    	s=fir=0;
	}
  printf("%d\n",2*solve(n>>1,(minn>>1)+1)%MOD);
  return 0;
}