You are given a string S of length n with each character being one of the first m lowercase English letters. 

Calculate how many different strings T of length n composed from the first m lowercase English letters exist such that the length of LCS (longest common subsequence) between S and T is n - 1.

Recall that LCS of two strings S and T is the longest string C such that C both in S and T as a subsequence.

Input

The first line contains two numbers n and m denoting the length of string S and number of first English lowercase characters forming the character set for strings (1 ≤ n ≤ 100 000, 2 ≤ m ≤ 26).

The second line contains string S.

Output

Print the only line containing the answer.

Examples

Input

3 3
aaa


Output

6


Input

3 3
aab


Output

11


Input

1 2
a


Output

1


Input

10 9
abacadefgh


Output

789

Note

For the first sample, the 6 possible strings T are: aab, aac, aba, aca, baa, caa. 

For the second sample, the 11 possible strings T are: aaa, aac, aba, abb, abc, aca, acb, baa, bab, caa, cab.

For the third sample, the only possible string T is b.
#include <bits/stdc++.h>
using namespace std;
int n, m;
long long dp[100010][8];
char s[100010];
int main() {
  scanf("%d%d%s", &n, &m, s + 1);
  for (int i = 1; i <= n; i++) s[i] -= 'a';
  for (int i = 0; i < m; i++) {
    int st = 0;
    if (i == s[1]) st ^= 2;
    if (i == s[1] || i == s[2]) st ^= 4;
    dp[1][st]++;
  }
  for (int i = 2; i <= n; i++) {
    for (int st = 0; st <= 7; st++) {
      int l[4] = {max((i - 3) + (st & 1), 0), (i - 2) + ((st >> 1) & 1),
                  (i - 2) + ((st >> 2) & 1)};
      for (int j = 0; j < m; j++) {
        int l2[4] = {0, 0, 0};
        for (int k = 1; k <= 3; k++) {
          if (s[i - 2 + k] == j)
            l2[k] = l[k - 1] + 1;
          else
            l2[k] = max(l2[k - 1], l[k]);
        }
        if (l2[1] < i - 2 || l2[2] < i - 1 || l2[3] < i - 1) continue;
        int nxt =
            (l2[1] - i + 2) ^ ((l2[2] - i + 1) << 1) ^ ((l2[3] - i + 1) << 2);
        dp[i][nxt] += dp[i - 1][st];
      }
    }
  }
  printf("%lld\n", dp[n][0] + dp[n][1] + dp[n][4] + dp[n][5]);
  return 0;
}
