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 bool OnLineJudge = 1;
const int N = 100005, oo = 1000 * 1000 * 1000 + 2604;
long long dp[N][4][4];
char a[N];
int n, m;
long long f(const int len, const int pos1, const int pos2) {
  if (len > n) return 1;
  if (dp[len][pos1][pos2] != -1) return dp[len][pos1][pos2];
  dp[len][pos1][pos2] = 0;
  for (char ch = 'a'; ch < 'a' + m; ch++) {
    int lenp, pos1p, pos2p;
    lenp = len + 1;
    pos1p = 3;
    if (pos1 != 3) {
      pos1p = 3;
      for (int i = -1; i < 2; i++)
        if (len + i <= n && len + i > len + pos1 - 2 && ch == a[len + i]) {
          pos1p = i + 1;
          break;
        }
    }
    pos2p = 3;
    for (int i = -1; i < 2; i++)
      if (len + i <= n && len + i > len + pos2 - 2 && ch == a[len + i]) {
        pos2p = i + 1;
        break;
      }
    if (pos1 != 3 && pos1 - 1 < pos2p) pos2p = pos1 - 1;
    if (pos2p == 3) continue;
    dp[len][pos1][pos2] += f(lenp, pos1p, pos2p);
  }
  return dp[len][pos1][pos2];
}
int main() {
  if (!OnLineJudge) {
    freopen("input.inp", "r", stdin);
  }
  std ::ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  memset(dp, -1, sizeof(dp));
  cout << f(1, 1, 1) - 1 << "\n";
  return 0;
}
