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;
int n, m;
char s[211111];
long long dp[211111][17];
int main() {
  scanf("%d %d\n", &n, &m);
  gets(s);
  dp[0][(1ll << (0))] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < (1ll << (4)); j++)
      if (dp[i][j]) {
        for (int c = 0; c < m; c++) {
          int can = 0;
          if (c == s[i] - 'a' && (j & (1ll << (0)))) can |= (1ll << (0));
          if (i + 1 < n && c == s[i + 1] - 'a' &&
              ((j & (1ll << (1))) || (j & (1ll << (0)))))
            can |= (1ll << (1));
          if (i > 0 && c == s[i - 1] - 'a' && (j & (1ll << (2))))
            can |= (1ll << (2));
          if (j & (1ll << (0))) can |= (1ll << (2));
          if (c == s[i] - 'a' && (j & (1ll << (3)))) can |= (1ll << (3));
          if (j & (1ll << (0))) can |= (1ll << (3));
          if (j & (1ll << (1))) can |= (1ll << (3));
          if (i > 0 && c == s[i - 1] - 'a' && (j & (1ll << (2))))
            can |= (1ll << (3));
          dp[i + 1][can] += dp[i][j];
        }
      }
  long long ans = 0;
  for (int i = 1; i < (1ll << (4)); i++)
    if ((i & (1ll << (0))) == 0 && (i & (1ll << (3)))) ans += dp[n][i];
  cout << ans << endl;
  return 0;
}
