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[100100];
long long dp[100100][2][2][2];
long long dfs(int x, int l1, int l2, int l3) {
  if (x == n) return l2 == n - 1;
  if (l1 < x - 2 || l2 < x - 1 || l3 < x - 1) return 0;
  long long& res = dp[x][l1 & 1][l2 & 1][l3 & 1];
  if (res != -1) return res;
  res = 0;
  int nl1, nl2, nl3;
  for (int i = 0; i < m; ++i) {
    nl1 = (i == s[x] - 'a' ? l1 + 1 : l2);
    nl2 = (i == s[x + 1] - 'a' ? l2 + 1 : max(nl1, l3));
    nl3 = (i == s[x + 2] - 'a' ? l3 + 1 : nl2);
    res += dfs(x + 1, nl1, nl2, nl3);
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  memset(dp, -1, sizeof(dp));
  printf("%lld\n", dfs(0, 0, 0, 0));
  return 0;
}
