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 <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <string>
#define SIZE 200005
#define MOD 1000000007

using namespace std;
typedef long long int ll;

char str[SIZE];
ll dp[SIZE],rdp[SIZE];

ll solve(int n,int L)
{
	dp[1]=1;
	rdp[1]=1;
	for(int i=2;i<=n;i++)
	{
		dp[i]=rdp[i-1]-(i-L-2<=0?0:rdp[i-L-2]);
		if(dp[i]<0) dp[i]+=MOD;
		rdp[i]=rdp[i-1]+dp[i];
		if(rdp[i]>=MOD) rdp[i]-=MOD;
	}
	ll ret=0;
	for(int i=0;i<=min(n-1,L);i++)
	{
		ret+=(ll) (i+1)*dp[n-i]%MOD;
		if(ret>=MOD) ret-=MOD;
	}
	return ret;
}
ll solve2(int n)
{
	dp[1]=1;
	rdp[1]=1;
	dp[2]=0;
	rdp[2]=1;
	for(int i=3;i<=n;i++)
	{
		dp[i]=rdp[i-2];
		rdp[i]=rdp[i-1]+dp[i];
		if(rdp[i]>=MOD) rdp[i]-=MOD;
	}
	ll ret=0;
	for(int i=1;i<=n-1;i++)
	{
		ret+=(ll) (i+1)*dp[n-i]%MOD;
		if(ret>=MOD) ret-=MOD;
	}
	return ret;
}
int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	scanf("%s",&str);
	bool R=false,B=false;
	for(int i=0;i<m;i++)
	{
		if(str[i]=='R') R=true;
		else B=true;
	}
	if(!B||!R) printf("%lld\n",(solve2(n)+1)%MOD);
	else
	{
		if(n%2==1) puts("0");
		else
		{
			int mn=n;
			for(int i=0;i<m;)
			{
				if(str[i]!=str[0])
				{
					i++;
					continue;
				}
				int f=i;
				for(;i<m&&str[i]==str[f];i++);
				int len=i-f;
				//printf("%d\n",len);
				if(f==0)
				{
					if(len%2==0) len++;
					mn=min(mn,len);
				}
				else if(i<m&&len%2==1) mn=min(mn,len);
			}
			//printf("%d\n",mn);
			printf("%lld\n",solve(n/2,mn/2)*2LL%MOD);
		}
	}
	return 0;
}
