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>
using namespace std;
typedef long long lint;
typedef pair<int, int> pi;
const int MAXN = 200005;
const int mod = 1e9 + 7;

int n, m;
int f[MAXN];
char s[MAXN];
int sum[MAXN];

int main(){
	scanf("%d %d %s",&n,&m,s);
	for(int i=0; i<m; i++) if(s[i] == 'R') s[i] = 'A';
	if(s[0] == 'B'){
		for(int i=0; i<m; i++) s[i] = 'A' + 'B' - s[i];
	}
	if(count(s, s + m, 'B') == 0){
		f[0] = 1; f[1] = 2;
		for(int i=2; i<=n; i++){
			f[i] = f[i-1] + f[i-2];
			f[i] %= mod;
		}
		lint ret = f[n];
		if(n == 3) ret += mod - 1;
		else if(n >= 4) ret += mod - f[n - 4];
		ret %= mod;
		cout << ret << endl;
		return 0;
	}
	int alt = 1e9;
	int pref = 0;
	while(s[pref] == 'A') pref++;
	for(int i=0; i<m; i++){
		if(s[i] == 'B'){
			int e = i + 1;
			while(s[e] == 'A') e++;
			if(e == m) break;
			if(e > i + 1 && e % 2 == i % 2){
				alt = min(alt, e - i - 1);
			}
			i = e - 1;
		}
	}
	if(n % 2 == 1){
		puts("0");
		return 0;
	}
	int x = min(alt, pref + 1);
	f[0] = 1;
	sum[0] = 1;
	for(int i=1; i<=n/2; i++){
		f[i] = sum[i - 1];
		if(i >= (x + 3) / 2) f[i] += mod - sum[i - (x + 3) / 2];
		f[i] %= mod;
		sum[i] = (sum[i-1] + f[i]) % mod;
	}
	lint ret = 0;
	for(int i=0; i<n;i++){
		int len = n - 1 - x;
		if(n - 1 - x > n - 1 - i) continue;
		ret += sum[(n - 1 - i) / 2];
		if(len >= 1){
			ret += mod - sum[(len - 1) / 2];
		}
	}

	cout << ret % mod << endl;
}
