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[2][9];
int nxt[100000 + 5][26];
char s[100000 + 5];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  for (int c = 0; c <= m - 1; c++) nxt[n + 1][c] = 4;
  for (int i = n; i >= 1; i--) {
    for (int c = 0; c <= m - 1; c++) nxt[i][c] = nxt[i + 1][c] + 1;
    nxt[i][s[i] - 'a'] = 0;
  }
  DP[(0) & 1][1 * 3] = 1;
  for (int i = 1; i <= n; i++) {
    for (int nmsk = 0; nmsk <= 9 - 1; nmsk++) DP[(i)&1][nmsk] = 0;
    for (int msk = 0; msk <= 9 - 1; msk++)
      if (DP[(i - 1) & 1][msk]) {
        int f0 = msk % 3, f1 = msk / 3;
        for (int c = 0; c <= m - 1; c++) {
          int g0 = min(2, f0 + nxt[i + f0][c]);
          int g1 = min(f0, f1 + nxt[i - 1 + f1][c]);
          int nmsk = g0 + g1 * 3;
          DP[(i)&1][nmsk] += DP[(i - 1) & 1][msk];
        }
      }
  }
  long long ans = -1;
  for (int msk = 0; msk <= 625 - 1; msk++)
    if (msk / 3 < 2) ans += DP[(n)&1][msk];
  printf("%lld\n", ans);
  return 0;
}
