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;

typedef long long ll;

const int MX = 200005;
const ll MOD = 1000000007;

template <typename T> void cmin(T &x, const T &y)
{
	if(y < x) x = y;
}

template <typename T> void cmax(T &x, const T &y)
{
	if(y > x) x = y;
}

template <typename T> void read(T &x)
{
	x = 0; char c = getchar(); bool f = 0;
	while(!isdigit(c) && c!='-') c = getchar();
	if(c == '-') f = 1, c = getchar();
	while(isdigit(c)) x = x*10+c-'0', c = getchar();
	if(f) x = -x;
}

int n, m, nR, nB;
char str[MX];
ll f[MX], sum[2][MX];

vector<int> divide(int x)
{
	vector<int> ret;
	for(int i=1; i<=x; i++)
		if(x%i == 0)
			ret.push_back(i);
	return ret;
}

int main()
{
	ll ans = 0;
	read(n), read(m);
	scanf("%s", str+1);
	for(int i=1; i<=m; i++)
		if(str[i] == 'R') nR++;
		else nB++;
	vector<int> dvs = divide(n);
	if(!nR || !nB)
	{
		f[0] = 1;
		sum[0][0] = sum[0][1] = 1;
		for(int i=2; i<=n; i++)
		{
			f[i] = sum[0][i-2];
			sum[0][i] = (sum[0][i-1] + f[i]) % MOD;
		}
		ans = 1;
		for(int i=1; i<n; i++) ans = (ans+f[n-i-1]*(i+1)) % MOD;
	}
	else
	{
		int len = n-1;
		int pre = 0;
		while(str[pre+1] == str[1]) pre++;
		/*if(pre&1) cmin(len, pre);		//the worst case isn't when the initial position is in the middle!
		else cmin(len, pre*2-1);*/
		cmin(len, pre|1);
		int cur = 0;
		for(int i=pre+1; i<=m; i++)
		{
			if(str[i] != str[1])
			{
				if(cur & 1) cmin(len, cur);
				cur = 0;
			}
			else cur++;
		}
		f[0] = sum[0][0] = 1;
		for(int i=1; i<=n; i++)
		{
			f[i] = (((i-2>=0) ? (sum[i&1][i-2]) : (0)) - ((i-2-len>=0) ? (sum[i&1][i-2-len]) : (0)) + MOD) % MOD;
			sum[i&1][i] = (sum[i&1][i-1] + f[i]) % MOD;
			sum[i&1^1][i] = sum[i&1^1][i-1];
		}
		for(int i=1; i<=len; i+=2) ans = (ans+f[n-i-1]*(i+1)) % MOD;
	}
	printf("%lld\n", ans);
	return 0;
}