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;

int n,m;
char str[200050];
const int mod = 1000000007;
int dp[200050];

int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",str + 1);
	bool c1 = 0,c2 = 0;
	for(int i = 1;i <= m; ++ i)
	{
		c1 |= str[i] == 'R';
		c2 |= str[i] == 'B';
	}
	int p = 1;
	int lim;
	if(c1 && c2)
	{
		if(n & 1)
		{
			printf("0\n");
			return 0;
		}
		for(;str[p + 1] == str[1]; ++ p);
		lim = p + (p % 2 == 0);
		int len = 0;
		for(int i = p + 1;i <= m; ++ i)
		{
			if(str[i] != str[1]) lim = min(lim,len % 2 ? len : 10000000),len = 0;
			else len ++;
		}
		lim ++;
		lim /= 2;
		lim = min(lim,n / 2);
		int pre = 1;
		int cur = 0;
		dp[0] = 1;
		long long ans = 0;
		for(int i = 0;i <= n / 2; ++ i)
		{
			if(i)
			{
				while(i - cur > lim) pre -= dp[cur],pre = (pre + mod) % mod,cur ++;
				dp[i] = pre;
				pre += dp[i];
				pre %= mod;
			}
			if(n - i * 2 > 0 && (n - i * 2) / 2 <= lim)
				ans = ans + 1LL * (n - i * 2) * dp[i] % mod,ans %= mod;
		}
		printf("%lld\n",ans);
	}
	else
	{
		int pre = 0;
		int cur = 0;
		dp[0] = 1;
		long long ans = 0;
		for(int i = 0;i <= n; ++ i)
		{
			if(i > 0)
			{
				dp[i] = pre;
				pre += dp[i - 1];
				pre %= mod;
			}
			if(n - i > 1)
				ans = ans + 1LL * (n - i) * dp[i] % mod,ans %= mod;
		}
		printf("%lld\n",(ans + 1) % mod);
	}
}