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;

const int N = 3e5 + 5, mod = 1e9 + 7;
int dp[N], sum[N];

char s[N];

int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  scanf("%s", s);
  bool same = 1;
  for (int i = 0; i < m; ++i)
    if (s[i] != s[0])
      same = 0;
  if (same) {
    dp[0] = 1;
    for (int i = 1; i <= n; ++i) {
      dp[i] = dp[i-1];
      if (i > 1)
        dp[i] = (dp[i] + dp[i-2]) % mod;
    }
    int ans = (dp[n] + dp[n-2]) % mod;
    printf("%d\n", ans);
    return 0;
  }
  int cnt = 0;
  while (s[cnt] == s[0]) ++cnt;
  if ((cnt & 1) == 0)
    ++cnt;
  int cur = 0;
  for (int i = 0; i < m; ++i) {
    if (s[i] != s[0]) {
      if (cur & 1) {
        cnt = min(cnt, cur);
      }
      cur = 0;
    }
    else {
      ++cur;
    }
  }
  dp[0] = 1;
  sum[0] = 1;
  for (int i = 2; i <= n; i += 2) {
    dp[i] = sum[i-2] - (i - (cnt+3) >= 0 ? sum[i - (cnt+3)] : 0);
    dp[i] %= mod;
    if (dp[i] < 0)
      dp[i] += mod;
    sum[i] = (dp[i] + sum[i-2]) % mod;
  }
  int ans = dp[n];
  for (int i = 2; i <= n && i <= cnt+1; i += 2) {
    ans = (ans + 1LL * dp[n-i] * (i-1)) % mod;
  }
  printf("%d\n", ans);
  return 0;
}
