#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 10;
string s;
long long n, m, cost;
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int lst = 0;
  cin >> n >> m >> s;
  --m;
  for (int i = 1; i < n; ++i) {
    if (s[i] == s[i - 1])
      cost += n * m, lst = i;
    else if (i > 1 && s[i] != s[i - 2])
      lst = i - 1;
    cost += i - lst;
  }
  cout << -cost + n * n * m << endl;
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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 INF = 1e9 + 10;
string s;
long long n, m, cost;
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int lst = 0;
  cin >> n >> m >> s;
  --m;
  for (int i = 1; i < n; ++i) {
    if (s[i] == s[i - 1])
      cost += n * m, lst = i;
    else if (i > 1 && s[i] != s[i - 2])
      lst = i - 1;
    cost += i - lst;
  }
  cout << -cost + n * n * m << endl;
  return 0;
}

```