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 N = 100005;
int n, m, p, q, s1[N], s2[N];
char s[N];
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  ans = n * (m - 1);
  for (int i = 1; i <= n - 1; i++) {
    s1[i] = s1[i - 1] + (s[i] != s[i + 1]);
    ans += s1[i] * (m - 1);
  }
  for (int i = n; i >= 2; i--) {
    s2[i] = s2[i + 1] + (s[i] != s[i - 1]);
    ans += s2[i] * (m - 1);
  }
  for (int i = 1; i <= n - 1; i++) {
    p = max(0, p - 2);
    while (i + p + 2 <= n && s[i + p] == s[i + p + 2]) p++;
    if (s[i] != s[i + 1]) ans -= p + 1;
    swap(p, q);
  }
  printf("%I64d\n", ans);
  return 0;
}
