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;
char st[100005];
int main() {
  int n;
  int m;
  scanf("%d %d", &n, &m);
  scanf("%s", st);
  vector<int> vt;
  int cnt = 1;
  for (int i = 1; i < n; i++) {
    if (st[i] == st[i - 1])
      cnt++;
    else {
      vt.push_back(cnt);
      cnt = 1;
    }
  }
  vt.push_back(cnt);
  long long ans = 1ll * vt.size() * (n * m - n);
  cnt = 0;
  for (int i = 1; i < n; i++) {
    if (st[i] != st[i - 1]) {
      if (i >= 2 && st[i - 2] == st[i])
        cnt++;
      else
        cnt = 2;
    } else {
      cnt = 1;
    }
    if (cnt) ans -= cnt - 1;
  }
  printf("%lld\n", ans);
  return 0;
}
