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 oo = 1e20;
const int MAX = 200000;
long long n, m;
char str[MAX + 5];
int p[MAX + 5];
int main() {
  scanf("%d", &n);
  scanf("%d", &m);
  scanf(" %s", &str);
  long long ans = n * (m - 1);
  for (int i = 1; i < n; i++) {
    if (str[i] != str[i - 1]) ans += i * (m - 1);
  }
  for (int i = 0; i < n - 1; i++) {
    if (str[i] != str[i + 1]) ans += (n - 1 - i) * (m - 1);
  }
  for (int i = 0; i < n; i++) {
    if (i < 2 || str[i] != str[i - 2])
      p[i] = 1;
    else
      p[i] = p[i - 2] + 1;
    if (i > 0) {
      int len = min(2 * p[i], 2 * p[i - 1] + 1);
      if (str[i] != str[i - 1] && len >= 2) ans -= len - 1;
    }
  }
  printf("%lld", ans);
  printf("\n");
  return 0;
}
