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 w[N];
char s[N];
long long g[N][8];
int rd() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return x;
}
int main() {
  int n = rd(), m = rd();
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) w[i] = s[i] - 96;
  g[0][7] = 1;
  for (int i = 0; i <= n - 1; i++)
    for (int j = 0; j < 8; j++) {
      if (!g[i][j]) continue;
      for (int k = 1; k <= m; k++) {
        int S = 0;
        if (k == w[i])
          S |= j & 1;
        else
          S |= (j >> 1) & 1;
        if (k == w[i + 1])
          S |= ((j >> 1) & 1) << 1;
        else if ((((j >> 2) | S) & 1) == 0)
          continue;
        if (k == w[i + 2])
          S |= ((j >> 2) & 1) << 2;
        else
          S |= ((S >> 1) & 1) << 2;
        g[i + 1][S] += g[i][j];
      }
    }
  printf("%I64d", g[n][0] + g[n][1]);
  return 0;
}
