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>

const int MOD = 1e9 + 7;

void update(int& x, int a)
{
  x += a;
  if (x >= MOD) {
    x -= MOD;
  }
}

int solve(int n, int m, const std::string& s)
{
  bool has_b = false;
  int max_len = -1;
  for (int i = 0, len = 0; i < m; ++ i) {
    if (i && s[i - 1] != s[i]) {
      if (s[i - 1] == s[0]) {
        if (max_len == -1) {
          max_len = len + 1;
        } else if (len & 1) {
          max_len = std::min(max_len, len + 1);
        }
      }
      // printf("%c %d\n", s[i - 1], len);
      len = 0;
    }
    has_b |= s[i] != s[0];
    len ++;
  }
  std::vector<int> dp(n);
  dp[0] = 1;
  if (has_b) {
    // <=> has A + has B + no consecutive B + len <= max_len
    for (int i = 2, j = 0, sum = 0; i < n; i += 2) {
      update(sum, dp[i - 2]);
      if (i - j > max_len + 1) {
        update(sum, MOD - dp[j]);
        j += 2;
      }
      dp[i] = sum;
    }
    int result = 0;
    for (int len = 2; len <= n && len - 1 <= max_len; len += 2) {
      update(result, 1LL * len * dp[n - len] % MOD);
    }
    return result;
  } else {
    // AA...A <=> has A + no consecutive B
    for (int i = 1, sum = 0; i < n; ++ i) {
      if (i >= 2) {
        update(sum, dp[i - 2]);
      }
      dp[i] = sum;
    }
    int result = 1; // no B
    for (int len = 2; len <= n; ++ len) {
      update(result, 1LL * len * dp[n - len] % MOD);
    }
    return result;
  }
}

int main()
{
  int n, m;
  while (std::cin >> n >> m) {
    std::string s;
    std::cin >> s;
    std::cout << solve(n, m, s) << std::endl;
  }
}
