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 <cstdio>
#include <algorithm>

typedef long long LL;
const int Mod = 1000000007;
const int MN = 200005, MM = 200005;

int N, M, K;
char S[MM];
int A[MN], B[MN], Ans;

int main() {
	scanf("%d%d%s", &N, &M, S + 1);
	K = 0x3f3f3f3f;
	int isfirst = 1, len = 0;
	for (int i = 1; i <= M; ++i) {
		if (S[i] != S[1]) {
			if (isfirst || len & 1) K = std::min(K, len / 2 + 1);
			isfirst = len = 0;
		} else ++len;
	}
	if (isfirst) {
		int X1 = 1, X2 = 0;
		for (int i = 1; i <= N; ++i) {
			X2 -= (X2 = X1 + X2) >= Mod ? Mod : 0;
			X1 += (X1 = X2 - X1) < 0 ? Mod : 0;
		}
		printf("%lld\n", (2ll * X1 + X2) % Mod);
		return 0;
	}
	if (N & 1) return puts("0"), 0;
	N /= 2;
	K = std::min(K, N);
	A[0] = B[0] = 1;
	for (int i = 1; i <= N; ++i) {
		A[i] = B[i - 1];
		if (i > K) A[i] += (A[i] -= B[i - K - 1]) < 0 ? Mod : 0;
		B[i] -= (B[i] = B[i - 1] + A[i]) >= Mod ? Mod : 0;
	}
	for (int i = 1; i <= K; ++i)
		Ans = (Ans + 2ll * i * A[N - i]) % Mod;
	printf("%d\n", Ans);
	return 0;
}
