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 = 2e5 + 10;
const int maxm = 2e5 + 10;
const long long mod = 1e9 + 7;
int getint() {
  char c;
  while ((c = getchar()) && !(c >= '0' && c <= '9') && c != '-')
    ;
  int ret = c - '0', sgn = 0;
  if (c == '-') sgn = 1, ret = 0;
  while ((c = getchar()) && c >= '0' && c <= '9') ret = ret * 10 + c - '0';
  if (sgn) ret = -ret;
  return ret;
}
char s[maxn];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) == 2) {
    scanf("%s", s);
    long long ans = (long long)n * (m - 1);
    for (int i = 1; i < n; i++) {
      if (s[i] != s[i - 1]) ans += (long long)n * (m - 1);
    }
    long long cur = 1;
    int i;
    for (i = 1; i <= n - 1; i++) {
      if (cur == 1) {
        if (s[i] == s[i - 1]) {
          continue;
        } else
          cur++;
      } else {
        if (s[i] == s[i - 2])
          cur++;
        else {
          ans -= (cur - 1) * cur / 2;
          if (s[i] == s[i - 1])
            cur = 1;
          else
            cur = 2;
        }
      }
    }
    ans -= cur * (cur - 1) / 2;
    cout << ans << endl;
  }
  return 0;
}
