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>
#define ll long long
using namespace std;
int Mn(int a,int b){return a<b?a:b;}
const int N=2e5+5,mod=1e9+7;
int upt(int x){while(x>=mod)x-=mod;while(x<0)x+=mod;return x;}

int n,m,lm,dp[N],sm[N],ans,g[N];
char s[N];
void solve()
{
  g[0]=sm[0]=1;
  for(int i=1;i<=n;i++)
    {
      if(i>=2)g[i]=sm[i-2];
      sm[i]=upt(sm[i-1]+g[i]);
    }
  int ans=1;//ans=1 for all one color
  for(int i=2;i<=n;i++)
    ans=(ans+(ll)i*g[n-i])%mod;
  printf("%d\n",ans);
}
int main()
{
  scanf("%d%d",&n,&m);
  scanf("%s",s+1); bool fg=0;
  for(int i=1;i<=m;i++)
    if(s[i]!=s[1]){fg=1;break;}
  if(!fg){solve();return 0;}
  if(n&1){puts("0");return 0;}
  lm=n+1; int lst;
  for(int i=m;i;i--)if(s[i]==s[1]){lst=i;break;}
  for(int i=1;i<=m;i++)
    if(s[i]==s[1])//== not !=
      {
	int j=i;
	while(j+1<=m&&s[j+1]==s[j])j++;
	swap(i,j); j=i-j+1;
	if(i-j==0) lm=Mn(lm,j+((j&1)==0));
	else if((j&1)&&i!=m)lm=Mn(lm,j);//i!=m
	//else if((j&1)&&i!=lst)lm=Mn(lm,j);
      }
  n>>=1; lm=(lm+1)>>1; dp[0]=sm[0]=1;
  for(int i=1;i<=n;i++)
    {
      dp[i]=sm[i-1]; if(i>lm)dp[i]=upt(dp[i]-sm[i-lm-1]);
      sm[i]=upt(sm[i-1]+dp[i]);
    }
  int ans=0;
  for(int i=1;i<=lm;i++)//lm not n
    ans=(ans+(ll)i*2*dp[n-i])%mod;
  printf("%d\n",ans);
  return 0;
}
