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>
long long D[101000], D2[101000], D3[101000][26], D4[101000][26], D5[101000][26],
    Res;
int m, n;
char p[101000];
int main() {
  int i, j, k;
  scanf("%d%d", &n, &m);
  scanf("%s", p + 1);
  for (i = 1; i <= n; i++) p[i] -= 'a';
  for (i = 1; i <= n; i++) {
    for (j = 0; j < m; j++)
      if (p[i] != j) D4[i][j] = 1;
    D[i] += D[i - 1];
    D2[i] += D2[i - 1];
    if (p[i] != p[i - 1]) D[i] += D2[i - 1];
    for (j = 0; j < m; j++) {
      if (j == p[i]) {
        for (k = 0; k < m; k++) D3[i][k] += D3[i - 1][j];
      } else {
        D[i] += D3[i - 1][j];
      }
    }
    for (j = 0; j < m; j++) {
      if (j == p[i]) {
        for (k = 0; k < m; k++) {
          if (k == p[i - 1])
            D5[i][k] += D4[i - 1][j];
          else
            D3[i][k] += D4[i - 1][j];
        }
        continue;
      }
      D2[i] += D4[i - 1][j];
      if (p[i] != p[i - 1]) D[i] += D4[i - 1][j];
    }
    for (j = 0; j < m; j++) {
      if (j == p[i]) {
        for (k = 0; k < m; k++) {
          if (k == p[i - 1])
            D5[i][k] += D5[i - 1][j];
          else
            D3[i][k] += D5[i - 1][j];
        }
        continue;
      }
      D2[i] += D5[i - 1][j];
      if (p[i] != p[i - 1]) D[i] += D5[i - 1][j];
    }
  }
  for (i = 0; i < m; i++) Res += D3[n][i] + D4[n][i] + D5[n][i];
  Res += D[n] + D2[n];
  printf("%lld\n", Res);
}
