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
//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 20021225
#define N 200100
#define mdn 1000000007
using namespace std;
int read()
{
	int s=0,t=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')	t=-1; ch=getchar();}
	while(ch>='0' && ch<='9')	s=s*10+ch-'0',ch=getchar();
	return s*t;
}
int f[N],n,m,pre[2][N]; char ch[N];
void upd(int &x,int y){x+=x+y>=mdn?y-mdn:y;}
int query(int l,int r,int w){return !l?pre[w][r]:(pre[w][r]-pre[w][l-1]+mdn)%mdn;}
int main()
{
	n=read(),m=read(); scanf("%s",ch+1);
	char w=ch[1]; int s=0,mn=inf,fs=0;
	for(int i=1;i<=m;i++)
		if(ch[i]==w)	s++;
		else	mn=s&1?min(s,mn):mn,fs=!fs?s:fs,s=0;
	mn=min(fs+!(fs&1),mn); f[0]=1,pre[0][1]=pre[0][0]=1;
	int ans=0;
	if(s==m)
	{
		f[0]=1; pre[0][0]=1;
		for(int i=2;i<=n;i++)
			f[i]=query(0,i-2,0),pre[0][i]=(pre[0][i-1]+f[i])%mdn;
		for(int i=1;i<=n;i++)	upd(ans,1ll*(i+1)*f[n-i-1]%mdn);
		printf("%d\n",ans+1);
		return 0;
	}
	for(int i=2;i<=n;i++)
		f[i]=query(max(0,i-mn-1),i-2,i&1),
		pre[i&1][i]=(pre[i&1][i-1]+f[i])%mdn,
		pre[!(i&1)][i]=pre[!(i&1)][i-1];
	for(int i=1;i<=mn;i+=2)	upd(ans,1ll*(i+1)*f[n-i-1]%mdn);
	printf("%d\n",ans);
	return 0;
}