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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod=1000000007;
inline void add(int &x,int y){(x+=y)>=mod?x-=mod:0;}
inline int kpow(int a,int b)
{
	int s=1;
	for(;b;b>>=1,a=1ll*a*a%mod)
		if(b&1)s=1ll*s*a%mod;
	return s;
}
int n,m,jc[200020],jcv[200020],ans,dp[200020];
char s[200020];
void init()
{
	jc[0]=1;
	for(int i=1;i<=200000;++i)jc[i]=1ll*jc[i-1]*i%mod;
	jcv[200000]=kpow(jc[200000],mod-2);
	for(int i=200000;i;--i)jcv[i-1]=1ll*jcv[i]*i%mod;
}
inline int C(int a,int b)
{
	if(a<b)return 0;
	return 1ll*jc[a]*jcv[b]%mod*jcv[a-b]%mod;
}
int main()
{
	init();
	scanf("%d %d",&n,&m);
	scanf("%s",s+1);
	char r=s[1];
	for(int i=1;i<=m;++i)s[i]=s[i]==r?0:1;
	int last=0,mi=0x3f3f3f3f;
	for(int i=1;i<=m;++i)
		if(s[i]==1)
		{
			if(!last)
			{
				if(i&1)mi=min(mi,i);
				else mi=min(mi,i-1);
			}
			else if((i-last-1)&1)mi=min(mi,i-last-1);
			last=i;
		}
	if(!last)
	{
		for(int k=0;k<=n-2-k;++k)add(ans,C(n-2-k,k));
		add(ans,ans);
		for(int k=0;k<=n-1-k;++k)add(ans,C(n-1-k,k));
		printf("%d",ans);
		return 0;
	}
	if(n&1)return printf("0"),0;
	n>>=1;mi=mi+1>>1;
	int tmp=1;dp[0]=1;
	for(int i=1;i<=n;++i)
	{
		if(i-mi-1>=0)add(tmp,mod-dp[i-mi-1]);
		dp[i]=tmp;
		add(tmp,dp[i]);
	}
	for(int i=1;i<=mi;++i)ans=(ans+1ll*dp[n-i]*i)%mod;
	add(ans,ans);
	printf("%d",ans);
	return 0;
}
