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 = 1<<19, mod = 1e9 + 7;
int n, m, pr = 0, dp[maxn], p[maxn][2];
string s;
int corner() {
	dp[0] = 2, dp[1] = 1;
	for(int i = 2; i <= n; i++) dp[i] = (dp[i-1] + dp[i-2])%mod;
	return dp[n];
}
void add(int &a, int b) {
	a += b;
	if(a >= mod) a -= mod;
}
int normal() {
	int ans = 0;
	for(int i = 0; i <= n; i++) {
		if(i) {
			add(p[i][0], p[i-1][0]);
			add(p[i][1], p[i-1][1]);
		}
		dp[i] = p[i][i&1];
		if(i == 1) {
			dp[i] = 1;
		}
		add(p[i+2][i&1], dp[i]);
		add(p[i+pr+2][i&1], mod - dp[i]);
	}
	for(int i = 1; i <= n; i++) {
		if(i + pr < n || ((i+pr)&1) != (n&1)) continue;
		ans = (ans + (n-i+1)*1ll*dp[i])%mod;
	}
	return ans;
}
bool check(int msk, int f = 0) {
	if((msk&1) && ((msk>>(n-1))&1)) return 0;
	if(msk == 0) return 0;
	int prev = -1, streak = 0, i = 0;
	if(f) while(!(msk&1)) {
		msk = (msk/2);}
	for(; i < n; i++) {
		int cur = (msk>>i)&1;
		if(prev == cur) streak++;
		else {
			if(prev == 1 && streak > 1) return 0;
			if(prev == 0 && (streak > pr || (streak&1) == 0)) return 0;
			streak = 1;
		}
		//cout << i << " " << cur << " " << streak << '\n';
		prev = cur;
	}
	if(prev == 1 && streak > 1) return 0;
			if(prev == 0 && (streak > pr || (streak&1) == 0)) return 0;
	
	return true;
}
int brute() {
	int ans = 0;
	for(int i = 0; i < 1<<n; i++) {
		ans += check(i, 1);
	}
	return ans;
}
int main() {
	cin >> n >> m >> s;
	if(s[0] == 'R')
		for(auto &i : s) i ^= 'B'^'R';
	while(pr < m && s[pr] == 'B') pr++;
	if(m == pr) return cout << corner(), 0;
	pr += pr%2 == 0;
	int cur = 0;
	for(auto &i : s) {
		if(i == 'B') cur++;
		else {
			if(cur&1) pr = min(cur, pr);
			cur = 0;
		}
	}
	cout << normal();
}
