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 MAX = 100005;
int pre[MAX], suf[MAX];
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  s = "#" + s;
  for (int i = n - 1; i >= 0; i--) {
    suf[i] = suf[i + 1];
    if (s[i] != s[i + 1]) suf[i]++;
  }
  for (int i = 2; i <= n; i++) {
    pre[i] = pre[i - 1];
    if (s[i] != s[i - 1]) pre[i]++;
  }
  long long ans = n * (m - 1);
  for (int i = 1; i < n; i++)
    if (s[i] != s[i + 1]) {
      ans += 2 * suf[i] * (m - 1);
      ans += (n - i - suf[i]) * (m - 1);
      ans += (i - pre[i + 1]) * (m - 1);
    }
  int l = 1;
  while (l < n) {
    if (s[l] == s[l + 1]) {
      l++;
      continue;
    }
    int r = l + 2;
    while (r <= n && s[r] == s[r - 2]) r++;
    int len = (r - l);
    ans -= 1LL * len * (len + 1) / 2;
    ans += len;
    l = r - 1;
  }
  cout << ans << endl;
  return 0;
}
