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 p = 1e9 + 7;
int main() {
  do {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  } while (false);
  int n, m;
  cin >> n >> m;
  string str;
  cin >> str;
  if (str[0] == 'B') {
    for (int i = 0; i < m; i++) {
      if (str[i] == 'R') {
        str[i] = 'B';
      } else {
        str[i] = 'R';
      }
    }
  }
  while (!str.empty() && str.back() == 'R') {
    str.pop_back();
  }
  if (str.empty()) {
    vector<int> dp(n);
    int sum = 0;
    for (int i = 1; i < n; i++) {
      dp[i] = (i + 1 + sum) % p;
      if (i) {
        sum = (sum + dp[i - 1]) % p;
      }
    }
    cout << (dp[n - 1] + 1) % p << endl;
  } else {
    int limit = INT_MAX, last = -1;
    for (int i = 0; i < str.size(); i++) {
      if (str[i] == 'B') {
        int len = i - last - 1;
        if (len & 1) {
          limit = min(limit, len);
        } else if (-1 == last) {
          limit = min(limit, len + 1);
        }
        last = i;
      }
    }
    vector<int> dp(n);
    int sum_0 = 0, sum_1 = 0;
    for (int i = 1; i < n; i++) {
      if (i & 1) {
        if (i <= limit) {
          dp[i] = i + 1;
        }
        dp[i] = (dp[i] + sum_1) % p;
        sum_0 = (sum_0 + dp[i - 1]) % p;
      } else {
        dp[i] = (dp[i] + sum_0) % p;
        sum_1 = (sum_1 + dp[i - 1]) % p;
      }
      if (i - limit - 1 > 0) {
        if (i - limit - 1 & 1) {
          sum_1 = (sum_1 - dp[i - limit - 1] + p) % p;
        } else {
          sum_0 = (sum_0 - dp[i - limit - 1] + p) % p;
        }
      }
    }
    cout << dp[n - 1] << endl;
  }
  return 0;
}
