#include <bits/stdc++.h>
using namespace std;
int main() {
  string S;
  int L, n, m, k = 1;
  cin >> n >> m >> S;
  L = S.length();
  for (int i = 0; i < L - 1; i++) {
    if (S[i] != S[i + 1]) {
      k++;
    }
  }
  long long ans = k * 1LL * (1LL * n * m - n);
  long long cnt = 1;
  for (int i = 1; i < L; i++) {
    if (S[i] != S[i - 1] && (i == 1 || S[i - 2] == S[i])) {
      cnt++;
    } else {
      ans -= cnt * (cnt - 1) / 2;
      if (S[i] == S[i - 1])
        cnt = 1;
      else
        cnt = 2;
    }
  }
  ans -= cnt * (cnt - 1) / 2;
  cout << ans << endl;
}

### Prompt

In cpp, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  string S;
  int L, n, m, k = 1;
  cin >> n >> m >> S;
  L = S.length();
  for (int i = 0; i < L - 1; i++) {
    if (S[i] != S[i + 1]) {
      k++;
    }
  }
  long long ans = k * 1LL * (1LL * n * m - n);
  long long cnt = 1;
  for (int i = 1; i < L; i++) {
    if (S[i] != S[i - 1] && (i == 1 || S[i - 2] == S[i])) {
      cnt++;
    } else {
      ans -= cnt * (cnt - 1) / 2;
      if (S[i] == S[i - 1])
        cnt = 1;
      else
        cnt = 2;
    }
  }
  ans -= cnt * (cnt - 1) / 2;
  cout << ans << endl;
}

```