#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e6 + 10;
const int MAXN = 1e6 + 10;
const int MOD = 1e9 + 7;
const int inf = 1e9;
const double pi = acos(-1.0);
const double eps = 1e-6;
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  long long res;
  string s;
  cin >> n >> m >> s;
  res = n * m - n;
  for (int i = 1; i < n; i++) res += (s[i] != s[i - 1]) * (n * m - n);
  long long cur = 1;
  for (int i = 1; i < n; i++) {
    if (cur == 1) {
      if (s[i] == s[i - 1])
        continue;
      else
        cur++;
    } else {
      if (s[i] == s[i - 2])
        cur++;
      else {
        res -= cur * (cur - 1) / 2;
        if (s[i] == s[i - 1])
          cur = 1;
        else
          cur = 2;
      }
    }
  }
  res -= cur * (cur - 1) / 2;
  cout << res;
  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 MAX = 1e6 + 10;
const int MAXN = 1e6 + 10;
const int MOD = 1e9 + 7;
const int inf = 1e9;
const double pi = acos(-1.0);
const double eps = 1e-6;
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  long long res;
  string s;
  cin >> n >> m >> s;
  res = n * m - n;
  for (int i = 1; i < n; i++) res += (s[i] != s[i - 1]) * (n * m - n);
  long long cur = 1;
  for (int i = 1; i < n; i++) {
    if (cur == 1) {
      if (s[i] == s[i - 1])
        continue;
      else
        cur++;
    } else {
      if (s[i] == s[i - 2])
        cur++;
      else {
        res -= cur * (cur - 1) / 2;
        if (s[i] == s[i - 1])
          cur = 1;
        else
          cur = 2;
      }
    }
  }
  res -= cur * (cur - 1) / 2;
  cout << res;
  return 0;
}

```