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 N = 100005;
int n, m;
char s[N];
long long cal1() {
  long long ret = 0;
  for (int i = 0; i + 1 < n; ++i) {
    if (s[i] != s[i + 1]) {
      ret += (long long)(m - 1) * (n - i - 1);
    }
  }
  return ret;
}
long long cal2() {
  long long ret = 0;
  for (int i = n - 1; i > 0; --i) {
    if (s[i] != s[i - 1]) {
      ret += (long long)(m - 1) * i;
    }
  }
  return ret;
}
long long cal3() {
  long long ret = 0, cur = 0, last = 0;
  for (int i = n - 1; i >= 0; --i) {
    if (i + 2 < n && s[i] != s[i + 1] && s[i] == s[i + 2]) {
      cur = last + 1;
    } else if (i + 1 < n && s[i] != s[i + 1]) {
      cur = 2;
    } else {
      cur = 1;
    }
    last = cur;
    ret += cur - 1;
  }
  return ret;
}
long long cal0() { return n * (m - 1); }
int main() {
  scanf("%d%d%s", &n, &m, s);
  cout << cal1() + cal2() - cal3() + cal0() << endl;
  return 0;
}
