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 mod=1000000007;
int n,m,f[200010],sum[200010],la,mi,ans,jc[200010],inv[200010];
char s[200010];
bool bo;

int quickmi(int a,int b)
{
	int t=1;
	for (; b; b>>=1,a=1ll*a*a%mod)
		if (b&1) t=1ll*t*a%mod;
	return t;
}

int calc(int n)
{
	int ans=0;
	for (int i=0; n-i-1>=i; i++)
		ans=(ans+1ll*jc[n-i-1]*inv[i]%mod*inv[n-2*i-1])%mod;
	return ans;
}

int main()
{
	scanf("%d%d",&n,&m),scanf("%s",s),bo=1;
	if (s[0]=='B') {for (int i=0; i<m; i++) s[i]=(s[i]=='B'?'R':'B');}
	for (int i=0; i<m; i++) if (s[i]=='B') {bo=0; break;}
	if (bo)
	{
		jc[0]=1;
		for (int i=1; i<=n; i++) jc[i]=1ll*jc[i-1]*i%mod;
		inv[n]=quickmi(jc[n],mod-2);
		for (int i=n; i; i--) inv[i-1]=1ll*inv[i]*i%mod;
		ans=(calc(n)+2ll*calc(n-1))%mod;
		return printf("%d\n",ans),0;
	}
	la=-1,mi=1000000000,ans=0;
	for (int i=0; i<m; i++)
		if (s[i]=='B') 
		{
			if ((i-la-1)&1) mi=min(mi,i-la-1);
			else if (la==-1) mi=min(mi,i-la);
			la=i;
		}
	f[0]=sum[0]=1,f[1]=sum[1]=0;
	for (int i=2; i<=n; i++) 
	{
		f[i]=sum[i-2]-(i-mi-3<0?0:sum[i-mi-3]);
		if (f[i]<0) f[i]+=mod;
		sum[i]=(sum[i-2]+f[i]>=mod?sum[i-2]+f[i]-mod:sum[i-2]+f[i]);
	}
	for (int i=0,nw1,nw2; i<=mi&&i<n; i++)
	{
		nw1=(n-i-1-((i&1)^1)>=0?sum[n-i-1-((i&1)^1)]:0);
		nw2=(n-i-3-(mi-i)>=0?sum[n-i-3-(mi-i)]:0);
		ans+=(nw1<nw2?nw1-nw2+mod:nw1-nw2);
		if (ans>=mod) ans-=mod;
	}
	printf("%d\n",ans);
	return 0;
}