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 = 200000 + 100;
int n, m;
char str[maxn];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    char c = getchar();
    while (c <= 32) c = getchar();
    str[i] = c;
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    cnt++;
    int j = i;
    while (j + 1 <= n && str[j + 1] == str[i]) j++;
    i = j;
  }
  long long ans = 1ll * cnt * n * (m - 1);
  for (int i = 1; i + 1 <= n; i++) {
    if (str[i] == str[i + 1]) continue;
    char s[2] = {str[i], str[i + 1]};
    int j = i + 1, k = 0;
    while (j + 1 <= n && str[j + 1] == s[k]) k ^= 1, j++;
    int len = j - i + 1;
    ans -= 1ll * len * (len - 1) / 2;
    i = j - 1;
  }
  cout << ans << endl;
  return 0;
}
