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>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair
 
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MOD = 1e9 + 7;
 
int n, m;
char S[200005];
vector<int> sizs;
int dp[200005], prs[200005];

int main()
{
	scanf("%d%d%s", &n, &m, S);
	
	int ccnt = 1;
	for(int i = 1; i < m; i ++)
	if(S[i] != S[i - 1]) {
		sizs.push_back(ccnt);
		ccnt = 1;
	} else ccnt ++;
	sizs.push_back(ccnt);
	
	if(sizs.size() == 1) {
		dp[1] = 2;
		dp[2] = MOD - 1;
		rep(i, n) {
			if(i >= 1) dp[i] = (dp[i] + dp[i - 1]) % MOD;
		}
		rep(i, n) {
			if(i >= 1) dp[i] = (dp[i] + dp[i - 1]) % MOD;
			if(i >= 2) dp[i] = (dp[i] + dp[i - 2]) % MOD;
		}
		printf("%d\n", dp[n - 1] + 1);
	} else {
		int maxl = min(n, sizs[0] + 2);
		rep(i, sizs.size()) if(i + 1 < sizs.size() && !(i & 1) && (sizs[i] & 1))
		maxl = min(maxl, sizs[i] + 1);
		
		rep1(i, maxl) if(!(i & 1)) dp[i] = i;
		
		rep1(i, n) {
			if(i >= 2) dp[i] = (dp[i] + prs[i - 2]) % MOD;
			if(i >= 2 + maxl) dp[i] = (dp[i] + (MOD - 1LL) * prs[i - 2 - maxl]) % MOD;
			if(i >= 2) prs[i] = (dp[i] + prs[i - 2]) % MOD;
		}
		
		printf("%d\n", dp[n]);
	}
	return 0;
}