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>
int n, m;
char s[100001];
int main() {
  scanf("%d %d", &n, &m);
  scanf("%s", s);
  long long c = 0;
  for (int i = 0; s[i] != '\0';) {
    int j = i;
    while (s[j] == s[i]) ++j;
    c += n * (m - 1);
    i = j;
  }
  for (int i = 0; s[i] != '\0';) {
    int j = i + 1;
    if (s[j] == '\0') break;
    if (s[j] == s[i]) {
      i = j;
    } else {
      int k = j + 1;
      while (s[k] == s[k - 2]) ++k;
      c -= static_cast<long long>(k - i) * (k - i - 1) / 2;
      i = k - 1;
    }
  }
  printf("%I64d\n", c);
}
