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 main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  long long k = 0, bad = 0, two = 1;
  for (int i = 0; i < n; i++) {
    if (!i || s[i - 1] != s[i]) k++;
    if (!i) continue;
    if (two == 1) {
      if (s[i - 1] != s[i]) two++;
    } else {
      if (s[i - 2] == s[i])
        two++;
      else {
        bad += ((two * (two - 1))) / 2;
        if (s[i - 1] != s[i])
          two = 2;
        else
          two = 1;
      }
    }
  }
  bad += ((two * (two - 1))) / 2;
  cout << k * (n * (m - 1)) - bad << "\n";
  return 0;
}
