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 inv(x) power(x,Mod-2)
#define NO() {puts("0");return 0;}
using namespace std;
typedef long long ll;
const int Mod=1e9+7;
int add(int a,const int b){a+=b;if(a>=Mod)a-=Mod;return a;}
int sub(int a,const int b){a-=b;if(a<0)a+=Mod;return a;}
int mul(const int a,const int b){return (ll)a*b%Mod;}
int power(int x,int cs)
{
	int ras=1;
	while(cs)
	{
		if(cs&1) ras=mul(ras,x);
		x=mul(x,x);cs>>=1;
	}
	return ras;
}
const int N=2e5+50;
int n,m,tag=0,mx,ans=0;
char s[N];
int val[N],dp[N],sum[N];
int q(int l,int r)
{
	if(l>r) return 0;
	int ras=sum[r];
	if(l!=0) ras=sub(ras,sum[l-1]);
	return ras;
}
int main()
{
	
	int i,x,pos;
	scanf("%d%d",&n,&m);	
	
	scanf("%s",s+1);
	if(s[1]=='B') tag=1;
	for(i=1;i<=m;i++)
	{
		if(s[i]=='R') val[i]=(1^tag);
		else val[i]=(0^tag);
	}
	for(x=1;x<=m&&val[x]==1;x++);
	x--;
	if(x==m)
	{
		mx=n;ans=1;
		dp[0]=1;sum[0]=1;
		for(i=1;i<=n;i++)
		{
			dp[i]=q(max(i-mx,0),i-2);
			sum[i]=add(sum[i-1],dp[i]);
		}
		//printf("%d %d\n",dp[1],dp[2]);
		for(i=2;i<=mx;i++)
		{
			ans=add(ans,mul(i,dp[n-i]));
		}
	}
	else
	{
		if(n&1) NO();
		if(x&1) mx=x;
		else mx=x+1;
		for(i=x+1;i<=m;)
		{
			if(val[i]==0) 
			{
				i++;continue;
			}
			pos=i;
			while(val[pos]==1) pos++;
			pos--;
			if(pos==m) break;
			if((pos-i+1)&1) mx=min(mx,pos-i+1);
			i=pos+1;
		}
		n=(n+1)/2;mx=(mx+1)/2;
		dp[0]=1;sum[0]=1;
		for(i=1;i<=n;i++)
		{
			dp[i]=q(max(i-mx,0),i-1);
			sum[i]=add(sum[i-1],dp[i]);
		}
		for(i=1;i<=mx;i++)
		{
			ans=add(ans,mul(i<<1,dp[n-i]));
		}
	}
	//printf("n%d mx%d\n",n,mx);
	
	
	printf("%d",ans);
	return 0;
}