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 + 5;
const int P = 1e9 + 7;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
int n, m;
char s[MAXN];
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
int main() {
	read(n), read(m);
	scanf("%s", s + 1);
	bool same = true;
	for (int i = 1; i <= m; i++)
		if (s[i] != s[1]) same = false;
	if (same) {
		static int dp[MAXN], sum[MAXN];
		dp[1] = sum[1] = 1;
		for (int i = 2; i <= n; i++) {
			dp[i] = sum[i - 2];
			sum[i] = (sum[i - 1] + dp[i]) % P;
		}
		int ans = (sum[n - 1] + 1) % P;
		for (int i = 2, j = n - 1; i <= n; i++, j--)
			update(ans, sum[j]);
		printf("%d\n", ans);
		return 0;
	}
	if (n & 1) {
		puts("0");
		return 0;
	}
	int Min = n - (n - 1) % 2;
	int last = 0;
	for (int i = 1; i <= m; i++)
		if (s[i] != s[1]) {
			int len = i - last - 1;
			if (last == 0) chkmin(Min, len + 1);
			if (len % 2 == 1) chkmin(Min, len);
			last = i;
		}
	static int dp[MAXN], sum[MAXN];
	dp[1] = sum[1] = 1;
	for (int i = 2; i <= n; i++) {
		if (i % 2 == 1) {
			dp[i] = sum[i - 2];
			update(dp[i], P - sum[max(i - Min - 3, 0)]);
		}
		sum[i] = (sum[i - 1] + dp[i]) % P;
	}
	int ans = (sum[n - 1] - sum[max(n - Min - 2, 0)] + P) % P;
	for (int i = 2, j = n - 1; i <= n && j >= n - Min - 2; i++, j--)
		update(ans, (sum[j] - sum[max(n - Min - 2, 0)] + P) % P);
	printf("%d\n", ans);
	return 0;
}