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>

typedef long long int int64;
static const int64 mod = 1000000007LL;

int n,m;
char s[200003];
int64 dp[200003],rdp[200003];
int64 solve(int n,int m) // divide circle of n into sections not longer than m
{
	dp[0] = rdp[0] = 1;
	for(int i = 1;i <= n;++i) {
		dp[i] = (rdp[i-1] - (i>=m+1?rdp[i-m-1]:0)) % mod;
		rdp[i] = (rdp[i-1] + dp[i]) % mod;
	}
	int64 ans = 0;
	for(int i = 1;i <= m;++i) {
		ans += (n-i>=0?rdp[n-i]:0) - (n-m-1>=0?rdp[n-m-1]:0);
	}
	return ans % mod;
}
int main()
{
	scanf("%d%d ",&n,&m);
	for(int i = 0;i < m;++i) {
		s[i] = getchar() == 'R' ? 0 : 1;
	}
	int len = 1;
	int maxlen = -1;
	for(int i = 1;i < m;++i) {
		s[i] ^= s[0];
		if(s[i] == 0) ++len;
		else {
			if(len == i) maxlen = (len%2==0?len+1:len);
			else if(len % 2 == 1) maxlen = std::min(maxlen,len);
			len = 0;
		}
	}
	int64 ans;
	if(len == m) {
		switch(n) {
		case 1: {
			ans = 1;
			break;
		}
		case 2: {
			ans = 3;
			break;
		}
		default: {
			dp[0] = dp[1] = 1;
			for(int i = 2;i <= n+1;++i) {
				dp[i] = (dp[i-1] + dp[i-2]) % mod;
			}
			ans = dp[n+1]-dp[n-3];
			break;
		}}
	} else {
		if(n%2==1) {
			ans = 0;
		} else {
			ans = solve(n/2,(maxlen+1)/2)*2;
		}
	}
	ans %= mod;
	if(ans < 0) ans += mod;
	printf("%lld\n", ans);
}
