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 maxn = 2e5, mod = 1e9 + 7;
int n, m, a[maxn + 10], lim;
char s[maxn + 10];
bool fd, fir = 1;
int f[maxn + 10], sum[maxn + 10], ans;
int main() {
	scanf("%d%d%s", &n, &m, s + 1);
	for (int i = 1; i <= m; ++i) a[i] = s[i] == 'R';
	if (a[1]) {
		for (int i = 1; i <= m; ++i) a[i] ^= 1;
	}
	for (int i = 1; i <= m; ++i)
		if (a[i]) fd = 1;
	lim = n;
	if (fd) {
		for (int i = 1, cnt = 0; i <= m; ++i)
			if (!a[i]) ++cnt;
			else {
				if (fir) {
					lim = min(lim, cnt % 2 ? cnt : cnt + 1);
					fir = 0;
				} else if (cnt % 2) 
					lim = min(lim, cnt);
				cnt = 0;
			}
		if (n % 2) {
			printf("0"); return 0;
		}
		n /= 2; lim = (lim + 1) / 2;
		f[0] = sum[0] = 1;
		for (int i = 1; i <= n; ++i) {
			if (i <= lim) f[i] = sum[i - 1];
			else f[i] = (sum[i - 1] - sum[i - lim - 1] + mod) % mod;
			sum[i] = (sum[i - 1] + f[i]) % mod;
		}
		for (int i = 1; i <= lim; ++i)
			(ans += 2ll * i * f[n - i] % mod) %= mod;
	} else {
		f[0] = sum[0] = 1;
		for (int i = 1; i <= n; ++i) {
			if (i > 1) f[i] = sum[i - 2];
			sum[i] = (sum[i - 1] + f[i]) % mod;
		}
		ans = 1;
		for (int i = 2; i <= lim; ++i)
			(ans += 1ll * i * f[n - i] % mod) %= mod;
	}
	printf("%d", ans);
}
