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 dp[100005];
int main() {
  long long n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  if (n == 1) {
    cout << m - 1 << endl;
    return 0;
  }
  long long ans = n * m - n;
  for (int i = 1; i < s.size(); i++)
    if (s[i] != s[i - 1]) ans += n * m - n;
  dp[n - 1] = 1;
  int len = (int)s.size();
  dp[len - 2] = (s[len - 2] == s[len - 1]) ? 1 : 2;
  for (int i = len - 3; i >= 0; i--) {
    if (s[i] == s[i + 1])
      dp[i] = 1;
    else if (s[i] == s[i + 2])
      dp[i] = dp[i + 1] + 1;
    else
      dp[i] = 2;
  }
  for (int i = 0; i < len; i++) ans -= dp[i] - 1;
  cout << ans << endl;
}
