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;

const int _ = 2e5 + 7 , MOD = 1e9 + 7;
char str[_]; int N , M;

int main(){
	scanf("%d %d %s" , &N , &M , str + 1);
	if(str[1] == 'B') for(int i = 1 ; i <= M ; ++i) str[i] = 'B' + 'R' - str[i];
	int limit = 1e9 , pre = 1;
	for(int i = 2 ; i <= M ; ++i)
		if(str[i] == 'B'){
			if(!pre) continue;
			if(pre == 1) limit = min(limit , (i - pre) / 2 * 2 + 1);
			if((i - pre) & 1) limit = min(limit , i - pre);
			pre = 0;
		}
		else if(!pre) pre = i;
	int ans = 0;
	if(limit == 1e9){
		static long long dp[_][2][2]; dp[2][0][0] = dp[2][1][0] = dp[2][0][1] = 1;
		for(int i = 3 ; i <= N ; ++i)
			for(int j = 0 ; j < 2 ; ++j){dp[i][j][0] = (dp[i - 1][j][0] + dp[i - 1][j][1]) % MOD; dp[i][j][1] = dp[i - 1][j][0];}
		ans = (dp[N][0][0] + dp[N][0][1] + dp[N][1][0]) % MOD;
	}
	else if(!(N & 1)){
		static int dp[_]; dp[1] = 1;
		if(limit >= N - 1) ans = N;
		for(int i = 3 ; i <= N ; i += 2){
			int val = (MOD + dp[i - 2] - dp[max(0 , i - limit - 3)]) % MOD;
			if(N - i <= limit) ans = (ans + 1ll * val * (N - i + 1)) % MOD;
			dp[i] = (dp[i - 2] + val) % MOD;
		}
	}
	cout << ans << endl; return 0;
}