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 ll long long
#define inf 1010000000
#define infll 1010000000000000000ll
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mpr make_pair
#define fi first
#define se second
#define vi vector<int>
#define pb push_back
#define mod 1000000007
#define ad(x,y) (x=(x+(y))%mod)
#define up(x,y) (x<(y)?x=(y):0)
#define dn(x,y) (x>(y)?x=(y):0)
#define N 200009
using namespace std;

int n,m,dp[N][2][2],f[N]; char ch[N]; vi e;
void solve0(){
	int i,j,k,l; dp[1][0][0]=dp[1][1][1]=1;
	for (i=2; i<=n; i++)
		for (j=0; j<2; j++)
			for (k=0; k<2; k++)
				for (l=0; k+l<2; l++)
					ad(dp[i][j][l],dp[i-1][j][k]);
	printf("%d\n",((dp[n][0][0]+dp[n][0][1])%mod+dp[n][1][0])%mod);
}
int main(){
	scanf("%d%d%s",&n,&m,ch+1);
	int i;
	if (ch[1]=='B')
		for (i=1; i<=m; i++)  ch[i]=(ch[i]=='R'?'B':'R');
	for (i=2; i<=m; i++) if (ch[i]=='B') break;
	if (i>m){
		solve0(); return 0;			
	}
	memset(dp,0,sizeof(dp));
	for (i=m; i; i--) if (ch[i]=='B') break;
	int lim=n,now=0;
	for (; i>=0; i--){
		if (!i || ch[i]=='B'){
			if (now) e.pb(now);
			now=0;
		} else now++;
	}
	reverse(e.begin(),e.end());
	for (i=0; i<e.size(); i++){
		if (i==0 && !(e[i]&1)) dn(lim,e[i]+2);
		if (e[i]&1) dn(lim,e[i]+1);	
	}
	f[0]=f[2]=1;
	for (i=3; i<=n; i++)
		f[i]=((ll)f[i-2]-(i-2-lim>=0?f[i-2-lim]:0)+f[i-2]+mod)%mod;
	int ans=0;
	for (i=2; i<=lim; i+=2) ad(ans,(ll)f[n-i]*i);
	printf("%d\n",ans);
	return 0;
}
