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 N = 200010;
const int mod = 1e9 + 7;

typedef long long LL;

char s[N];

int f[N], g[N], dp[2][2][N];

int main() {
    int n, m; scanf("%d%d%s", &n, &m, s + 1);
    bool flag = false;
    for (int i = 1; i <= m; i++) if (s[i] != s[1]) {
        flag = true; break;
    } 
    if (flag) {
        if (n & 1) return puts("0"), 0;
        int fir = 0;
        for (int i = 1; i <= m; i++) if (s[i] != s[1]) {
            fir = i - 1; break;
        }
        int lim = fir | 1;
        for (int i = fir + 1; i <= m; i++) if (s[i] != s[1]) {
            int j = i + 1;
            while (j <= m && s[j] == s[1]) j++;
            if (j <= m && (j - i - 1) % 2 == 1) lim = min(lim, j - i - 1); 
        } 
        n /= 2, lim = (lim + 1) / 2;
        for (int i = 1; i <= lim; i++) f[i] = i * 2;
        for (int i = 1, diff = 0; i <= n; i++) {
            diff = (diff + g[i]) % mod, f[i] = (f[i] + diff) % mod;
            if (i + 1 <= n) g[i + 1] = (g[i + 1] + f[i]) % mod;
            if (i + lim + 1 <= n) g[i + lim + 1] = (g[i + lim + 1] - f[i] + mod) % mod;
        }
        printf("%d\n", f[n]);
    } else {
        dp[1][1][1] = dp[0][0][1] = 1;
        for (int i = 2; i <= n; i++) 
            for (int t = 0; t <= 1; t++) {
                dp[t][0][i] = (dp[t][1][i - 1] + dp[t][0][i - 1]) % mod;
                dp[t][1][i] = dp[t][0][i - 1];
            }
        printf("%d\n", ((LL)dp[1][0][n] + dp[0][0][n] + dp[0][1][n]) % mod);
    }
    return 0;
}//