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;
const int MAXN = 100100;
int n, m;
char a[MAXN];
int main() {
  scanf("%d%d%s", &n, &m, a);
  long long ans = 0;
  char last = 'z' + 1;
  for (int i = 0; i < n; ++i) {
    if (a[i] != last) {
      ans += (long long)n * m - n;
    }
    last = a[i];
  }
  int sub_length = 1;
  for (int i = 1; i < n + 1; ++i) {
    if (sub_length == 1) {
      sub_length += a[i] != a[i - 1];
    } else {
      if (a[i] == a[i - 2]) {
        ++sub_length;
      } else {
        ans -= (long long)sub_length * (sub_length - 1) / 2;
        sub_length = a[i] == a[i - 1] ? 1 : 2;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}
