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;
char s[112345];
int n, m;
long long gao() {
  int ss = 0;
  long long ans = 0;
  for (int i = 0; i < int(n); ++i) {
    ss += m - 1;
    if (i + 1 < n && s[i + 1] != s[i]) {
      ans += ss;
    }
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s);
  long long ans = n * (m - 1);
  ans += gao();
  reverse(s, s + n);
  ans += gao();
  for (int i = 0; i + 1 < n; ++i) {
    char x[2] = {s[i], s[i + 1]};
    if (x[0] == x[1]) continue;
    int j;
    for (j = i + 1; j < n && s[j] == x[(j - i) & 1]; ++j)
      ;
    int len = j - i;
    ans -= len * ((long long)len + 1) / 2 - len;
    i = j - 2;
  }
  printf("%lld\n", ans);
}
