#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
char s[N];
int n, m;
long long num() {
  long long res = 0;
  for (int i = 1; i < n; ++i)
    if (s[i] != s[i + 1]) res += (m - 1) * (n - i);
  return res;
}
int main() {
  scanf("%d%d\n", &n, &m);
  scanf("%s", s + 1);
  long long res = num();
  for (int i = 1; i < n + 1 - i; ++i) swap(s[i], s[n + 1 - i]);
  res += num();
  long long seq = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i - 1] == s[i + 1] && s[i] != s[i - 1])
      ++seq;
    else if (s[i - 1] != s[i + 1]) {
      if (s[i] != s[i - 1]) ++seq;
      res -= seq * (seq - 1) / 2;
      seq = 0;
    }
  }
  for (int i = 1; i < n; ++i)
    if (s[i] != s[i + 1]) --res;
  printf("%I64d", res + n * (m - 1));
  return 0;
}

### Prompt

Please formulate 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 N = 1e5 + 5;
char s[N];
int n, m;
long long num() {
  long long res = 0;
  for (int i = 1; i < n; ++i)
    if (s[i] != s[i + 1]) res += (m - 1) * (n - i);
  return res;
}
int main() {
  scanf("%d%d\n", &n, &m);
  scanf("%s", s + 1);
  long long res = num();
  for (int i = 1; i < n + 1 - i; ++i) swap(s[i], s[n + 1 - i]);
  res += num();
  long long seq = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i - 1] == s[i + 1] && s[i] != s[i - 1])
      ++seq;
    else if (s[i - 1] != s[i + 1]) {
      if (s[i] != s[i - 1]) ++seq;
      res -= seq * (seq - 1) / 2;
      seq = 0;
    }
  }
  for (int i = 1; i < n; ++i)
    if (s[i] != s[i + 1]) --res;
  printf("%I64d", res + n * (m - 1));
  return 0;
}

```