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 maxn = 100005;
char c[maxn];
int main() {
  int i, n, m;
  cin >> n >> m;
  scanf("%s", c);
  long long s = 0;
  for (i = 0; i < n; i++) {
    int p = i;
    while (p < n - 1 && c[p] == c[p + 1]) p++;
    s += n * (m - 1);
    i = p;
  }
  for (i = 0; i < n - 1; i++) {
    if (c[i] == c[i + 1]) continue;
    int p = i;
    int q = i + 1;
    while (p < n - 2 && c[p] == c[p + 2]) {
      q = p + 2;
      p++;
    }
    int w = q - i + 1;
    s -= (long long)w * (w - 1) / 2;
    i = q - 1;
  }
  cout << s << endl;
}
