#include <bits/stdc++.h>
using namespace std;
int n, m;
char a[100000];
int main() {
  cin >> n >> m;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
  }
  long long rv = n * m - n;
  for (int i = 1; i < n; i++) {
    if (a[i] != a[i - 1]) {
      rv += n * m - n;
    }
  }
  long long c = 1;
  for (int i = 1; i < n; i++) {
    if (c == 1) {
      if (a[i] == a[i - 1]) {
        continue;
      } else {
        c++;
      }
    } else {
      if (a[i] == a[i - 2]) {
        c++;
      } else {
        rv -= (c * (c - 1)) / 2;
        if (a[i] == a[i - 1]) {
          c = 1;
        } else {
          c = 2;
        }
      }
    }
  }
  cout << rv - (c * (c - 1)) / 2 << 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;
int n, m;
char a[100000];
int main() {
  cin >> n >> m;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
  }
  long long rv = n * m - n;
  for (int i = 1; i < n; i++) {
    if (a[i] != a[i - 1]) {
      rv += n * m - n;
    }
  }
  long long c = 1;
  for (int i = 1; i < n; i++) {
    if (c == 1) {
      if (a[i] == a[i - 1]) {
        continue;
      } else {
        c++;
      }
    } else {
      if (a[i] == a[i - 2]) {
        c++;
      } else {
        rv -= (c * (c - 1)) / 2;
        if (a[i] == a[i - 1]) {
          c = 1;
        } else {
          c = 2;
        }
      }
    }
  }
  cout << rv - (c * (c - 1)) / 2 << endl;
  return 0;
}

```