#include <bits/stdc++.h>
using namespace std;
int read() {
  char c = getchar();
  int x = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x;
}
const int N = 1e5 + 10;
char s[N];
int main() {
  int n = read(), m = read();
  scanf("%s", s + 1);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i] != s[i - 1]) ans += 1ll * n * (m - 1);
  }
  int len = 1;
  for (int i = 2; i <= n; ++i) {
    if (len == 1) {
      if (s[i] == s[i - 1])
        continue;
      else
        len++;
    } else if (s[i] == s[i - 2]) {
      len++;
    } else {
      ans -= 1ll * len * (len - 1) / 2;
      if (s[i] == s[i - 1])
        len = 1;
      else
        len = 2;
    }
  }
  ans -= 1ll * len * (len - 1) / 2;
  cout << ans << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 read() {
  char c = getchar();
  int x = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x;
}
const int N = 1e5 + 10;
char s[N];
int main() {
  int n = read(), m = read();
  scanf("%s", s + 1);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i] != s[i - 1]) ans += 1ll * n * (m - 1);
  }
  int len = 1;
  for (int i = 2; i <= n; ++i) {
    if (len == 1) {
      if (s[i] == s[i - 1])
        continue;
      else
        len++;
    } else if (s[i] == s[i - 2]) {
      len++;
    } else {
      ans -= 1ll * len * (len - 1) / 2;
      if (s[i] == s[i - 1])
        len = 1;
      else
        len = 2;
    }
  }
  ans -= 1ll * len * (len - 1) / 2;
  cout << ans << endl;
  return 0;
}

```