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 = 1e5 + 10;
int n, m, a[N], p, q[5], nq[5];
long long f[N][10];
char s[N];
int main() {
  scanf("%d%d%s", &n, &m, s + 1);
  for (int i = 1; i <= n; i++) a[i] = s[i] - 'a' + 1;
  for (int i = 1; i <= m; i++) {
    p = 1;
    if (i == a[1]) p |= 2;
    if (i == a[1] || i == a[2]) p |= 4;
    ++f[1][p];
  }
  for (int i = 1; i <= n - 1; i++) {
    for (int j = 0; j <= 7; j++) {
      if (f[i][j]) {
        q[0] = i - 2 + (j & 1);
        q[1] = i - 1 + (j >> 1 & 1);
        q[2] = i - 1 + (j >> 2 & 1);
        for (int k = 1; k <= m; k++) {
          int p = 0;
          nq[0] = (k == a[i] ? q[0] + 1 : q[1]);
          nq[1] = (k == a[i + 1] ? q[1] + 1 : max(nq[0], q[2]));
          nq[2] = (k == a[i + 2] ? q[2] + 1 : nq[1]);
          if (nq[0] < i - 1 || nq[1] < i || nq[2] < i) continue;
          p = nq[0] - (i - 1);
          p |= (nq[1] - i) << 1;
          p |= (nq[2] - i) << 2;
          f[i + 1][p] += f[i][j];
        }
      }
    }
  }
  printf("%lld\n", f[n][0] + f[n][1] + f[n][4] + f[n][5]);
  return 0;
}
