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;
long long dp[100000 + 5][8];
char a[100000 + 5];
int main() {
  int n, m;
  scanf("%d%d%s", &n, &m, a + 1);
  for (int i = 1; i <= n; i++) a[i] -= 'a';
  for (int c = 0; c <= m - 1; c++)
    dp[1][1 | ((a[1] == c) << 1) | ((a[1] == c || a[2] == c) << 2)]++;
  for (int i = 1; i <= n - 1; i++)
    for (int c = 0; c <= m - 1; c++)
      for (int msk = 0; msk <= 7; msk++)
        if (dp[i][msk]) {
          int nmsk = 0;
          if (a[i] == c)
            nmsk |= msk & 1;
          else if (msk & 2)
            nmsk |= 1;
          if (a[i + 1] == c)
            nmsk |= msk & 2;
          else if ((msk & 4) == 0 && (nmsk & 1) == 0)
            continue;
          if (a[i + 2] == c)
            nmsk |= msk & 4;
          else if (nmsk & 2)
            nmsk |= 4;
          dp[i + 1][nmsk] += dp[i][msk];
        }
  long long ans = 0;
  for (int msk = 0; msk <= 7; msk++)
    if ((msk & 2) == 0) ans += dp[n][msk];
  printf("%lld\n", ans);
  return 0;
}
