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;
long long f[100010][8];
char ch[100010];
void work() {
  scanf("%d %d\n", &n, &m);
  scanf("%s", ch + 1);
  for (int i = 1; i <= n; i++) ch[i] -= 'a' - 1;
  for (int i = 1; i <= m; i++) {
    int x = 1;
    if (i == ch[1]) x |= 2;
    if ((i == ch[2]) || (i == ch[1])) x |= 4;
    f[1][x]++;
  }
  for (int i = 2; i <= n; i++)
    for (int j = 0; j < 8; j++) {
      int g[4] = {i - 3 + ((j & 1) > 0), i - 2 + ((j & 2) > 0),
                  i - 2 + ((j & 4) > 0), 0};
      for (int k = 1; k <= m; k++) {
        int h[4] = {0, 0, 0, 0};
        for (int l = 1; l <= 3; l++)
          if (ch[i + l - 2] == k)
            h[l] = g[l - 1] + 1;
          else
            h[l] = max(h[l - 1], g[l]);
        if ((h[1] >= i - 2) && (h[2] >= i - 1) && (h[3] >= i - 1))
          f[i][(h[1] == i - 1) | ((h[2] == i) << 1) | ((h[3] == i) << 2)] +=
              f[i - 1][j];
      }
    }
  cout << f[n][0] + f[n][1] + f[n][2] + f[n][3];
}
int main() {
  work();
  return 0;
}
