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>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
char s[100100];
int l[100100], f[100100];
int main() {
  int i, n, p;
  long long a = 0;
  scanf("%d%d%s", &n, &p, s);
  for (i = n - 1; i >= 0; i--) {
    f[i] = f[i + 1] + (s[i] != s[i + 1]);
    l[i] = 1;
    if (s[i] == s[i + 2]) l[i] += l[i + 2];
    if (i == n - 1 || s[i] == s[i + 1])
      a += (p - 1) * f[i];
    else
      a += (p - 2) * f[i] + (long long)(n - i - 2) * (p - 1) + p + f[i + 2] -
           min(2 * l[i], 2 * l[i + 1] + 1) + 2 - (s[i + 1] == s[i + 2]);
  }
  printf("%lld\n", a);
  return 0;
}
