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 long long NR = 1e5 + 10;
void Min(long long& x, long long y) { x = min(x, y); }
void Max(long long& x, long long y) { x = max(x, y); }
long long n, m;
long long ans;
char a[NR];
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
signed main() {
  n = read(), m = read();
  for (long long i = 1; i <= n; i++) scanf(" %c", &a[i]);
  for (long long i = 1; i <= n; i++) ans += n * (m - 1) * (a[i] != a[i - 1]);
  long long now = 0;
  for (long long i = 2; i <= n; i++) {
    if (!now)
      now += (a[i] != a[i - 1]);
    else if (a[i] == a[i - 2])
      now++;
    else if (a[i] != a[i - 1])
      ans -= (now + 1) * now / 2, now = 1;
    else
      ans -= (now + 1) * now / 2, now = 0;
  }
  ans -= now * (now + 1) / 2;
  printf("%lld\n", ans);
  return 0;
}
