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;
long long n, m, ans;
string st;
int main() {
  cin >> n >> m;
  cin >> st;
  ans = n * m - n;
  for (int k = 0; k < st.size(); ++k) {
    if (k > 0 && st[k] != st[k - 1]) ans += n * m - n;
  }
  long long c = 1;
  for (int k = 1; k < st.size(); ++k) {
    if (c == 1) {
      if (st[k] == st[k - 1]) continue;
      if (st[k] != st[k - 1]) {
        ++c;
      }
      continue;
    }
    if (st[k] == st[k - 2])
      ++c;
    else {
      ans -= (c * (c - 1)) / 2;
      if (st[k] != st[k - 1])
        c = 2;
      else
        c = 1;
    }
  }
  ans -= (c * (c - 1)) / 2;
  cout << ans;
  return 0;
}
