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 <iostream>
#include <string>
#include <vector>
#include <utility>
#include <cassert>
using namespace std;
typedef long long ll;
ll dp[200010] = {},fib[200010],mod = 1000000007;
string s;
vector<pair<char,int>> v;
int main(){
	ll i,j,n,m;
	cin >> n >> m >> s;
	if(s[0]=='B'){
		for(i=0;i<m;i++){
			if(s[i]=='R') s[i] = 'B';
			else s[i] = 'R';
		}
	}
	for(i=0;i<m;i++){
		if(i==0 || v.back().first!=s[i]){
			v.push_back({s[i],1});
		}else{
			v.back().second++;
		}
	}
	if(v.size()==1){
		fib[0] = 1,fib[1] = 1;
		for(i=2;i<=n;i++){
			fib[i] = (fib[i - 1] + fib[i - 2])%mod;
		}
		cout << (fib[n] + fib[n - 2])%mod << endl;
		return 0;
	}
	if(n&1){
		cout << 0 << endl;
		return 0;
	}
	if(v.back().first=='R'){
		v.pop_back();
	}
	int mn = -1;
	for(i=0;i<v.size();i++){
		if(i==0){
			mn = v[i].second + 1 - (v[i].second&1);
		}
		if(v[i].first=='R' && (v[i].second&1)){
			mn = min(mn,v[i].second);
		}
	}
	mn++;
	dp[0] = 1;
	ll sum = 1;
	for(i=2;i<=n;i+=2){
		(dp[i] += sum) %= mod;
		(sum += dp[i]) %= mod;
		if(i>=mn) sum -= dp[i - mn];
		if(sum<0) sum += mod;
	}
	ll ans = 0;
	for(i=2;i<=mn;i+=2){
		(ans += i*dp[n - i]) %= mod;
	}
	cout << ans << endl;
}