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;
#define nn 500008
#define mod 1000000007
char str[nn];int n,m;
long long dp[nn],sum[nn];
#define err return 0*puts("0");
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",str+1);
	
	int cnt=0;
	if(str[1]=='B') for(int i=1;i<=m;i++) if(str[i]=='R') str[i]='B';else str[i]='R';
	for(int i=1;i<=m;i++) if(str[i]=='B') cnt++;
	int mx=n;int flg=0;
	if(cnt)
	{
		cnt=0;
		if(n%2) err;int j=1;
		for(int i=m;i>=1;i--)
			if(str[i]=='B')  {for(int j=i;j<=m;j++) str[j]='B';break;}

		for(int i=1;i<=m;i++)
		{
			if(str[i]=='B') break;
			cnt++;j=i;
		}
		mx=min(mx,cnt/2+1);
		
		j++;
		cnt=0;str[m+1]='B';
		for(;j<=m+1;j++)
		{
			if(str[j]=='B')
			{
				if(cnt%2) mx=min(mx,cnt/2+1);
				cnt=0;
			}
			else cnt++;
		}
		flg=1;
		n/=2;
	}
	
	dp[0]=sum[0]=1;
	int ls=0;if(!flg) ls=1,sum[1]=1;
	for(int i=ls+1;i<=n;i++) 
	{
		if(i-mx-1>=0 and flg) dp[i]=(sum[i-1-ls]-sum[i-mx-1]+mod)%mod;
		else dp[i]=sum[i-1-ls];
		sum[i]=(sum[i-1]+dp[i])%mod;
	}
	long long ans=0;
	for(int i=1+ls;i<=n and i<=mx;i++)
	{
		ans+=1ll*dp[n-i]*i%mod;
		if(flg) ans+=1ll*dp[n-i]*i%mod;
		ans%=mod;
	}
	if(ls==1) ans++;
	
	printf("%lld\n",(ans%mod+mod)%mod);
	return 0;
}