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 maxn = 100200;
char s[maxn];
int n, m;
long long ans;
long long calc(int len) { return 1ll * (1 + len - 1) * (len - 1) / 2; }
int main() {
  scanf("%d%d%s", &n, &m, s + 1);
  for (int i = (1), iend = (n - 1); i <= iend; i++) {
    if (s[i] != s[i + 1]) {
      ans += n - i;
    }
  }
  for (int i = (2), iend = (n); i <= iend; i++) {
    if (s[i] != s[i - 1]) {
      ans += i - 1;
    }
  }
  ans += n;
  ans *= m - 1;
  for (int i = (1), iend = (n - 1); i <= iend; i++) {
    char bk[2];
    bk[1] = s[i], bk[0] = s[i + 1];
    if (bk[1] == bk[0]) continue;
    int cnt = 1, now = i;
    while (now < n && bk[((cnt % 2) ^ 1)] == s[now + 1]) {
      now++;
      cnt++;
    }
    i = now - 1;
    ans -= calc(cnt);
  }
  cout << ans;
  return 0;
}
