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>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 +c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,M,L;
char s[MAXN];
bool flag = 0;
int f[MAXN][2],dp[MAXN],sum[MAXN][2];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
void Solve1() {
    int ans = 0;
    f[1][1] = 1;f[1][0] = 1;
    for(int i = 2 ; i <= N ; ++i) {
	f[i][0] = f[i - 1][1];
	f[i][1] = inc(f[i - 1][1],f[i - 1][0]);
    }
    ans = inc(f[N][0],f[N][1]);
    memset(f,0,sizeof(f));
    f[1][0] = 1;
    for(int i = 2 ; i < N ; ++i) {
	f[i][0] = f[i - 1][1];
	f[i][1] = inc(f[i - 1][1],f[i - 1][0]);
    }
    ans = inc(ans,MOD - f[N - 1][1]);
    out(ans);enter;
}
void Solve2() {
    dp[0] = 1;sum[0][0] = 1;
    for(int i = 1 ; i <= N ; ++i) {
	if(i >= 2) {
	    int t = sum[i - 2][i & 1];
	    if(i > L + 1) update(t,MOD - sum[i - L - 2][i & 1]);
	    dp[i] = t;
	}
	sum[i][0] = sum[i - 1][0];sum[i][1] = sum[i - 1][1];
	sum[i][i & 1] = inc(sum[i][i & 1],dp[i]);
	
    }
    int ans = dp[N];
    for(int i = 1 ; i <= L ; i += 2) {
	update(ans,mul(i,dp[N - i - 1]));
    }
    out(ans);enter;
}
void Init() {
    read(N);read(M);
    scanf("%s",s + 1);
    while(M >= 1 && s[M] == s[1]) --M;
    L = N;
    if(M == 0) flag = 1;
    else {
	for(int i = 1 ; i <= M ; ++i) {
	    if(s[i] != s[1]) continue;
	    else {
		int j = i;
		while(s[j + 1] == s[1]) ++j;
		int t = j - i + 1;
		if(i == 1) {
		    if(t % 2 == 0) L = min(L,t + 1);
		    else L = min(L,t); 
		}
		else {
		    if(t % 2 == 1) L = min(L,t);
		}
		i = j;
	    }
	}
    }
    if(L == N) flag = 1;
    if(flag) Solve1();
    else Solve2();
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
}
