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 P = 1000000007;
int n, m, dp[200005], sum[2][200005];
char s[200005];

int main() {
	scanf("%d%d%s", &n, &m, s + 1);
	if (count(s + 1, s + 1 + m, s[1]) == m) {
		if (n == 1) return puts("1"), 0;
		int ans = 0;
		for (int c = 0; c < 2; ++c) {
			int dp[2] = {};
			dp[c] = 1;
			for (int i = 2; i <= n; ++i) {
				int t = dp[1];
				dp[1] = dp[0];
				dp[0] = (dp[0] + t) % P;
			}
			ans = (ans + dp[0]) % P;
			if (!c) ans = (ans + dp[1]) % P;
		}
		printf("%d\n", ans);
		return 0;
	}
	int lim = n - 1;
	for (int i = 1, j; i <= m; i = j) {
		for (j = i; j <= m && s[j] == s[i]; ++j);
		if (j > m || s[i] != s[1]) continue;
		if (j - i & 1)
			lim = min(lim, j - i);
		else if (i == 1)
			lim = min(lim, j - i + 1);
	}
	dp[0] = 1;
	sum[0][0] = sum[0][1] = 1;
	sum[1][0] = sum[1][1] = 0;
	for (int i = 2; i <= n; ++i) {
		int d = max(0, i - lim - 1);
		dp[i] = (sum[i & 1][i - 2] - (d == 0 ? 0 : sum[i & 1][d - 1])) % P;
		sum[0][i] = sum[0][i - 1];
		sum[1][i] = sum[1][i - 1];
		sum[i & 1][i] = (sum[i & 1][i] + dp[i]) % P;
	}
	long long ans = 0;
	for (int i = 0; i <= lim; ++i)
		ans += sum[n & 1][n - (i + 1)] - (n == lim + 1 ? 0 : sum[n & 1][n - (lim + 1) - 1]);
	printf("%lld\n", (ans % P + P) % P);
	return 0;
}
