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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<string>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
const int N = 2e5+10 , mod = 1e9+7;
inline int read()
{
	register int x = 0 , f = 0; register char c = getchar();
	while(c < '0' || c > '9') f |= c == '-' , c = getchar();
	while(c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0' , c = getchar();
	return f ? -x : x;
}
int n , m;
char s[N];
int sum[N] , f[N];
int main()
{
	n = read(); m = read(); scanf("%s" , s+1);
	int i , mx , ans = 0;
	for(i = 1 ; i <= m ; ++i) if(s[i] != s[1]) break;
	mx = i - 1;
	if(mx == m)
	{
		sum[0] = f[0] = ans = 1;
		for(int i = 1 ; i <= n ; ++i)
		{
			if(i >= 2) f[i] = sum[i-2];
			sum[i] = (sum[i-1] + f[i]) % mod;
		}
		for(int i = 0 ; n - i >= 2 ; ++i) (ans += (LL)f[i] * (n - i) % mod) %= mod;
		cout << ans << '\n'; return 0;
	}
	if(n & 1) return puts("0") , 0;
	if(mx % 2 == 0) mx++;
	for(int tmp = 0 ; i <= m ; ++i)
	{
		if(s[i] == s[1]) tmp++;
		else
		{
			if(tmp & 1) mx = min(mx , tmp);
			tmp = 0;
		}
	}
	mx = (mx + 1) >> 1; n >>= 1; f[0] = sum[0] = 1;
	for(int i = 1 ; i <= n ; ++i)
	{
		f[i] = (sum[i-1] - (i - mx - 1 >= 0 ? sum[i - mx - 1] : 0) + mod) % mod;
		sum[i] = (sum[i-1] + f[i]) % mod;
	}
	for(int i = 0 ; i <= n ; ++i) if(n - i <= mx) (ans += (LL)f[i] * (n - i) % mod * 2 % mod) %= mod;
	cout << ans << '\n'; return 0;
}