#include <bits/stdc++.h>
using namespace std;
const int N = 100100;
const unsigned long long M = 1LL << 33;
const double eps = 0;
char s[N];
int main() {
  int n, m;
  cin >> n >> m;
  scanf("%s", s);
  long long res = 0;
  for (int i = 0; i < n; i++) {
    if (i == 0 || s[i] != s[i - 1]) {
      res += n * (m - 1);
    }
  }
  int e = 0;
  for (int i = 0; i < n - 1; i++) {
    if (s[i] == s[i + 1]) continue;
    e = max(e, i + 1);
    while (e < n - 1 && s[e + 1] == s[e - 1]) e++;
    res -= max(0, e - i);
  }
  cout << res << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 = 100100;
const unsigned long long M = 1LL << 33;
const double eps = 0;
char s[N];
int main() {
  int n, m;
  cin >> n >> m;
  scanf("%s", s);
  long long res = 0;
  for (int i = 0; i < n; i++) {
    if (i == 0 || s[i] != s[i - 1]) {
      res += n * (m - 1);
    }
  }
  int e = 0;
  for (int i = 0; i < n - 1; i++) {
    if (s[i] == s[i + 1]) continue;
    e = max(e, i + 1);
    while (e < n - 1 && s[e + 1] == s[e - 1]) e++;
    res -= max(0, e - i);
  }
  cout << res << endl;
}

```