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 = (int)2e5 + 10;
char str[N];
int jump[N];
int pos[N];
int cntP = 0;
int getCountPos(int l, int r) {
  return upper_bound(pos, pos + cntP, r) - lower_bound(pos, pos + cntP, l);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  scanf(" %s", str);
  long long ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (i >= n - 2) jump[i] = i;
    if (str[i] == str[i + 2])
      jump[i] = jump[i + 2];
    else
      jump[i] = i + 2;
  }
  for (int i = n - 1; i > 0; i--) {
    if (str[i] != str[i - 1]) pos[cntP++] = i;
  }
  sort(pos, pos + cntP);
  for (int i = 0; i < n; i++) {
    ans += m - 1;
    if (i == n - 1) break;
    ans += (long long)getCountPos(i + 1, n) * (m - 1);
    if (str[i] != str[i + 1]) {
      ans += (long long)(m - 1) * (n - i - 1);
      int near = min(jump[i], jump[i + 1]);
      ans -= getCountPos(i + 1, near - 1);
    }
  }
  cout << ans << endl;
  return 0;
}
