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 ll;
const int N = 100054, mod = 1000000007;

int n, m, L;
int f[N];
char s[N];

inline void down(int &x, const int y) {x > y ? x = y : 0;}
inline void add(int &x, const int y) {x += y - mod, x += x >> 31 & mod;}
inline void sub(int &x, const int y) {x -= y, x += x >> 31 & mod;}

int main() {
	int i, j, u, v, ans = 0;
	scanf("%d%d%s", &L, &n, s);
	for (i = 0; i < n; ++i) s[i] = s[i] >> 4 & 1;
	if (*s) for (i = 0; i < n; ++i) s[i] ^= 1;
	for (j = 0; j < n && !s[j]; ++j);
	if (j == n) {
		for (u = 2, v = i = 1; i < L; ++i) j = u, u = v, add(v, j);
		return printf("%d\n", v), 0;
	}
	if (L & 1) return putchar(48), putchar(10), 0;
	m = j | 1, L /= 2;
	for (i = j + 1; i < n; ++i) if (s[i]) (i ^ j) & 1 || (down(m, i - j - 1), 0), j = i;
	assert(m & 1), m = (m + 1) / 2;
	for (f[1] = *f = i = 1; i < L; ++i)
		if (add(f[i + 1] = f[i], f[i]), i >= m) sub(f[i + 1], f[i - m]);
	for (i = std::max(L - m, 0); i < L; ++i) ans = (ans + ll(L - i) * f[i]) % mod;
	printf("%d\n", ans * 2 % mod);
	return 0;
}