#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
char s[maxn];
int main() {
  int n, m;
  while (~scanf("%d %d", &n, &m)) {
    scanf("%s", s);
    long long cnt = 1;
    for (int i = 0; i < n; ++i) {
      if (i == 0 || s[i] == s[i - 1]) continue;
      cnt++;
    }
    long long ans = cnt * (n * m - n);
    cnt = 1;
    for (int i = 1; i < n; ++i) {
      if (cnt == 1) {
        if (s[i] != s[i - 1]) cnt++;
      } else {
        if (s[i] == s[i - 2]) {
          cnt++;
        } else {
          ans -= (cnt * (cnt - 1)) / 2;
          if (s[i] == s[i - 1])
            cnt = 1;
          else
            cnt = 2;
        }
      }
    }
    ans -= (cnt * (cnt - 1)) / 2;
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Your task is to create 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 maxn = 1e5 + 10;
char s[maxn];
int main() {
  int n, m;
  while (~scanf("%d %d", &n, &m)) {
    scanf("%s", s);
    long long cnt = 1;
    for (int i = 0; i < n; ++i) {
      if (i == 0 || s[i] == s[i - 1]) continue;
      cnt++;
    }
    long long ans = cnt * (n * m - n);
    cnt = 1;
    for (int i = 1; i < n; ++i) {
      if (cnt == 1) {
        if (s[i] != s[i - 1]) cnt++;
      } else {
        if (s[i] == s[i - 2]) {
          cnt++;
        } else {
          ans -= (cnt * (cnt - 1)) / 2;
          if (s[i] == s[i - 1])
            cnt = 1;
          else
            cnt = 2;
        }
      }
    }
    ans -= (cnt * (cnt - 1)) / 2;
    printf("%I64d\n", ans);
  }
  return 0;
}

```