#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long n, m, k, cc;
string s;
int main() {
  k = 1;
  cc = 0;
  cin >> n >> m >> s;
  for (int i = 1; i < (int)n; i++) {
    if (s[i] != s[i - 1]) k++;
  }
  for (int i = 0; i < (int)n - 1; i++) {
    if (s[i] != s[i + 1]) {
      long long j = i;
      for (; i + 2 < n; i++) {
        if (s[i + 2] != s[i]) break;
      }
      j = (i - j + 2);
      cc += j * (j - 1) / 2;
    }
  }
  cout << k * (m * n - n) - cc << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long n, m, k, cc;
string s;
int main() {
  k = 1;
  cc = 0;
  cin >> n >> m >> s;
  for (int i = 1; i < (int)n; i++) {
    if (s[i] != s[i - 1]) k++;
  }
  for (int i = 0; i < (int)n - 1; i++) {
    if (s[i] != s[i + 1]) {
      long long j = i;
      for (; i + 2 < n; i++) {
        if (s[i + 2] != s[i]) break;
      }
      j = (i - j + 2);
      cc += j * (j - 1) / 2;
    }
  }
  cout << k * (m * n - n) - cc << endl;
  return 0;
}

```