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>
const int maxn = 1e5 + 5;
char s[maxn];
int cnt[26], n, m;
int main() {
  scanf("%d%d%s", &n, &m, s + 1);
  long long ans = 0;
  for (int i = 2; i <= n; i++)
    if (s[i] != s[i - 1]) ans += n * (m - 1);
  int cnt = 1;
  ans += n * (m - 1);
  for (int i = 2; i <= n; i++) {
    if (cnt == 1 && s[i] != s[i - 1] || cnt > 1 && s[i] == s[i - 2])
      cnt++;
    else
      ans -= 1ll * (cnt - 1) * cnt / 2, cnt = s[i] == s[i - 1] ? 1 : 2;
  }
  ans -= 1ll * (cnt - 1) * cnt / 2;
  printf("%lld\n", ans);
  return 0;
}
