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;
const int MAXN = 100009;
int n, m;
char s[MAXN];
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s), ans = n * m - n;
  for (int i = (1); i <= (n - 1); ++i)
    if (s[i] != s[i - 1]) ans += n * m - n;
  long long len = 1;
  for (int i = (1); i <= (n - 1); ++i)
    if (len == 1) {
      if (s[i] == s[i - 1])
        continue;
      else
        len++;
    } else {
      if (s[i] == s[i - 2])
        len++;
      else {
        ans -= (len * (len - 1)) >> 1;
        if (s[i] != s[i - 1])
          len = 2;
        else
          len = 1;
      }
    }
  ans -= (len * (len - 1)) >> 1;
  cout << ans << endl;
  return 0;
}
