#include <bits/stdc++.h>
using namespace std;
char s[100005];
int n, m;
int main(void) {
  scanf("%d%d%s", &n, &m, s);
  long long sol = (long long)n * m - n;
  for (int i = 1; i < n; ++i)
    sol += (s[i] != s[i - 1]) * ((long long)n * m - n);
  int curr = 1;
  for (int i = 1; i < n; ++i) {
    if (curr == 1) {
      if (s[i] == s[i - 1]) continue;
      ++curr;
    } else {
      if (s[i] == s[i - 2]) {
        ++curr;
      } else {
        sol -= (long long)curr * (curr - 1) / 2;
        curr = 1 + (s[i] != s[i - 1]);
        ;
      }
    }
  }
  sol -= (long long)curr * (curr - 1) / 2;
  printf("%lld\n", sol);
  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;
char s[100005];
int n, m;
int main(void) {
  scanf("%d%d%s", &n, &m, s);
  long long sol = (long long)n * m - n;
  for (int i = 1; i < n; ++i)
    sol += (s[i] != s[i - 1]) * ((long long)n * m - n);
  int curr = 1;
  for (int i = 1; i < n; ++i) {
    if (curr == 1) {
      if (s[i] == s[i - 1]) continue;
      ++curr;
    } else {
      if (s[i] == s[i - 2]) {
        ++curr;
      } else {
        sol -= (long long)curr * (curr - 1) / 2;
        curr = 1 + (s[i] != s[i - 1]);
        ;
      }
    }
  }
  sol -= (long long)curr * (curr - 1) / 2;
  printf("%lld\n", sol);
  return 0;
}

```