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 = 100010;
int n, m, k;
long long ans;
char ch[N];
int main() {
  scanf("%d%d", &n, &m);
  cin >> ch + 1;
  for (int i = 2; i <= n + 1; i++)
    if (ch[i] != ch[i - 1])
      ans = ans + (long long)n * (long long)m - (long long)n;
  k = 1;
  for (int i = 2; i <= n + 1; i++) {
    if (k == 1 && ch[i] != ch[i - 1])
      k = 2;
    else if (k == 1)
      continue;
    else if (ch[i] == ch[i - 2])
      k++;
    else {
      ans -= (long long)k * (long long)(k - 1) / 2;
      if (ch[i] == ch[i - 1])
        k = 1;
      else
        k = 2;
    }
  }
  printf("%lld", ans);
  getchar();
  getchar();
  return 0;
}
