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>
#pragma warning(disable : 4996)
using namespace std;
const int MN = 100000;
char buf[MN + 4];
vector<pair<int, int> > mtlr;
int mtcnt[MN];
int rcnt[MN];
int lcnt[MN];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  scanf("%s", buf);
  long long ans = (long long)n * n * m;
  int avk = 0;
  int curl = 0;
  for (int i = 0; i < n; i++) {
    if (i == 0 || buf[i] == buf[i - 1]) {
      mtcnt[avk]++;
    } else {
      avk++;
      mtcnt[avk]++;
      mtlr.push_back(make_pair(curl, i - 1));
      curl = i;
    }
  }
  mtlr.push_back(make_pair(curl, n - 1));
  avk++;
  ans -= ((long long)n * n + (long long)(avk - 1) * n -
          (long long)(avk) * (avk - 1));
  ans -= (long long)(n - avk) * ((m - 1) * n - (avk - 1));
  int i = 0;
  while (i < n - 1) {
    int x = buf[i];
    int y = buf[i + 1];
    if (x == y) {
      i++;
      continue;
    }
    int cnt = 0;
    while (buf[i + cnt] + buf[i + cnt + 1] == x + y) {
      cnt++;
    }
    ans -= (long long)cnt * (cnt + 1) / 2;
    i = i + cnt;
  }
  printf("%I64d", ans);
  return 0;
}
