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(void) {
  int64_t n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  vector<pair<int64_t, char>> blocks;
  int64_t blocklen = 1;
  for (int64_t i = 1; i < s.size(); i++) {
    if (s[i] == s[i - 1]) {
      blocklen++;
    } else {
      blocks.push_back(make_pair(blocklen, s[i - 1]));
      blocklen = 1;
    }
  }
  blocks.push_back(make_pair(blocklen, s[s.size() - 1]));
  int64_t total = 0;
  int64_t before = 0;
  total = blocks.size() * n * (m - 1);
  if (s.size() == 1) {
    cout << m - 1 << endl;
    return 0;
  }
  int64_t run = 0;
  if (s[0] != s[1]) {
    run = 1;
  }
  for (int64_t i = 2; i < s.size(); i++) {
    if (s[i] == s[i - 1]) {
      total -= run * (run + 1) / 2;
      run = 0;
    } else if (s[i] == s[i - 2]) {
      run++;
    } else {
      total -= run * (run + 1) / 2;
      run = 1;
    }
  }
  total -= run * (run + 1) / 2;
  cout << total << endl;
  return 0;
}
