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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_set>
#include <bitset>
#include <time.h>
#include <limits.h>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define Fi first
#define Se second
#define pb push_back
#define szz(x) (int)x.size()
#define rep(i,n) for(int i=0;i<n;i++)
#define all(x) x.begin(),x.end()
typedef tuple<int, int, int> t3;

const int mod = 1e9 + 7;

int n, m;
char S[200020];
ll res[200020];
ll D[200020], sum[200020];

int main() {
	scanf("%d%d", &n, &m);
	scanf("%s", S + 1);
	if(S[1] == 'B') {
		for(int i=1;i<=m;i++) S[i] = (S[i] == 'B' ? 'R' : 'B');
	}
	int all_r = 1;
	for(int i=1;i<=m;i++) if(S[i] != 'R') all_r = 0;
	if(all_r) {
		int dp[200020][2] = {};
		dp[1][0] = 1; dp[1][1] = 0;
		for(int i=2;i<=n;i++) {
			dp[i][0] = (dp[i-1][0] + dp[i-1][1]) % mod;
			dp[i][1] = dp[i-1][0];
		}
		int ans = (dp[n][0] + dp[n][1]) % mod;
		memset(dp, 0, sizeof dp);
		dp[1][0] = 0; dp[1][1] = 1;
		for(int i=2;i<=n;i++) {
			dp[i][0] = (dp[i-1][0] + dp[i-1][1]) % mod;
			dp[i][1] = dp[i-1][0];
		}
		ans = (ans + dp[n][0]) % mod;
		printf("%d\n", ans);
		return 0;
	}
	while(S[m] == 'R') --m;
	vector <int> v;
	for(int i=1, c=0;i<=m;i++) {
		if(S[i] == 'R') ++c;
		if(S[i] == 'R' && (i == m || S[i+1] == 'B')) {
			v.pb(c);
			c = 0;
		}
	}
	
	if(n % 2) {
		puts("0");
		return 0;
	}
	int lmn = n - 1;
	if(v[0] % 2) lmn = min(lmn, v[0]);
	else lmn = min(lmn, v[0] + 1);
	for(int i=1;i<szz(v);i++) {
		if(v[i] % 2) {
			lmn = min(lmn, v[i]);
		}
	}
	lmn = (lmn + 1) / 2;
	
	int sn = n / 2;
	
	D[0] = sum[0] = 1;
	for(int i=1;i<=sn;i++) {
		D[i] = (sum[i-1] - (i-lmn-1 < 0 ? 0 : sum[i-lmn-1]));
		if(D[i] < 0) D[i] += mod;
		sum[i] = (sum[i-1] + D[i]) % mod;
	}
	
	ll ans = 0;
	for(int i=1;i<=lmn;i++) {
		ans = (ans + (2 * i) * D[sn - i]) % mod;
	}
	printf("%lld\n", ans);
	
	return 0;
}
