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 <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
typedef long double ld;
#define mp make_pair
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll MOD = (ll)1e9 + 7;
ll add(ll x, ll y) {
	x += y;
	if (x >= MOD) return x - MOD;
	return x;
}
ll sub(ll x, ll y) {
	x -= y;
	if (x < 0) return x + MOD;
	return x;
}
ll mult(ll x, ll y) {
	return (x * y) % MOD;
}

const int N = 200200;
char s[N];
int n, m;
ll dp[N];

void solve1() {
	dp[0] = 0;
	dp[1] = 1;
	for (int i = 1; i <= n + 1; i++) {
		dp[i] = add(dp[i], dp[i - 1]);
		dp[i + 1] = dp[i - 1];
	}
	ll ans = 1;
	for (int i = 2; i <= n; i++)
		ans = add(ans, mult(i, sub(dp[n - i + 1], dp[n - i])));
	printf("%lld\n", ans);
	return;
}

int main()
{
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d%d", &n, &m);
	scanf("%s", s);
	bool allEq = true;
	for (int i = 1; i < m; i++)
		allEq &= s[i] == s[0];
	if (allEq) {
		solve1();
		return 0;
	}
	if (n & 1) {
		printf("0\n");
		return 0;
	}
	int b = n;
	int lst = -1;
	for (int i = 0; i < m; i++) {
		if (s[i] == s[0]) continue;
		int x = i - lst - 1;
		if (x % 2 == 0) {
			if (lst == -1) b = min(b, x + 2);
		} else {
			b = min(b, x + 1);
		}
		lst = i;
	}
	b /= 2;
	n /= 2;

	dp[0] = 0;
	dp[1] = 1;
	for (int i = 1; i <= n + 1; i++) {
		dp[i] = add(dp[i], dp[i - 1]);
		dp[i + 1] = sub(dp[i], dp[max(0, i - b)]);
	}
	ll ans = 0;
	for (int i = 1; i <= b; i++)
		ans = add(ans, mult(2 * i, sub(dp[n - i + 1], dp[n - i])));
	printf("%lld\n", ans);

	return 0;
}
