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;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int n, m;
  string s;
  cin >> n >> m >> s;
  unordered_map<long long, long long> dp;
  auto at = [&](int i, int a, int b, int c) {
    a -= i - 5;
    b -= i - 5;
    c -= i - 5;
    return i + n * 1LL * (a + 5 * (b + 5 * c));
  };
  function<long long(int, int, int, int)> f = [&](int i, int a, int b,
                                                  int c) -> long long {
    if (i == n) return (b == n - 1);
    if (b < i - 1) return 0;
    long long key = at(i, a, b, c);
    if (dp.count(key)) return dp[key];
    long long ans = 0;
    for (char ti = 'a'; ti < 'a' + m; ++ti) {
      int aa = b;
      if (i > 0 && ti == s[i - 1]) aa = a + 1;
      int bb = max(aa, c);
      if (ti == s[i]) bb = b + 1;
      int cc = bb;
      if (i + 1 < n && ti == s[i + 1]) cc = c + 1;
      ans += f(i + 1, aa, bb, cc);
    }
    return dp[key] = ans;
  };
  long long ans = f(0, 0, 0, 0);
  cout << ans << endl;
}
