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;
inline long long read() {
  long long flag = 1;
  long long x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x * flag;
}
const long long N = 1e5 + 10;
long long ans = 0;
char str[N];
long long len, m;
signed main() {
  len = read();
  m = read();
  scanf("%s", str + 1);
  for (long long i = 1; i <= len; i++)
    ans += len * (m - 1) * (str[i] != str[i - 1]);
  long long res = 1;
  for (long long i = 2; i <= len; i++) {
    if (res == 1)
      res += (str[i - 1] != str[i]);
    else if (str[i - 2] == str[i])
      res++;
    else {
      ans -= (res * (res - 1)) >> 1;
      res = (str[i] == str[i - 1]) ? 1 : 2;
    }
  }
  ans -= (res * (res - 1)) >> 1;
  printf("%lld\n", ans);
  return 0;
}
