#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
long long n, m, ans;
char s[N];
int main() {
  cin >> n >> m;
  cin >> s + 1;
  s[0] = '/';
  ans = 1ll * n * (m - 1);
  long long len = 1;
  for (int i = 2; i <= n; i++) {
    if (s[i] != s[i - 1]) {
      ans += n * (m - 1);
      if (len == 1) {
        len++;
        continue;
      }
    }
    if (len != 1) {
      if (s[i] == s[i - 2]) {
        len++;
      } else {
        ans -= len * (len - 1) / 2;
        if (s[i] == s[i - 1]) {
          len = 1;
        } else {
          len = 2;
        }
      }
    }
  }
  if (len > 1) ans -= 1ll * len * (len - 1) / 2;
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
const int N = 1000005;
long long n, m, ans;
char s[N];
int main() {
  cin >> n >> m;
  cin >> s + 1;
  s[0] = '/';
  ans = 1ll * n * (m - 1);
  long long len = 1;
  for (int i = 2; i <= n; i++) {
    if (s[i] != s[i - 1]) {
      ans += n * (m - 1);
      if (len == 1) {
        len++;
        continue;
      }
    }
    if (len != 1) {
      if (s[i] == s[i - 2]) {
        len++;
      } else {
        ans -= len * (len - 1) / 2;
        if (s[i] == s[i - 1]) {
          len = 1;
        } else {
          len = 2;
        }
      }
    }
  }
  if (len > 1) ans -= 1ll * len * (len - 1) / 2;
  cout << ans << endl;
  return 0;
}

```