Problem:
Score : 400 points

Problem Statement

Given are integers A, B, and N.

Find the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N.

Here floor(t) denotes the greatest integer not greater than the real number t.

Constraints

1 ≤ A ≤ 10^{6}

1 ≤ B ≤ 10^{12}

1 ≤ N ≤ 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B N

Output

Print the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N, as an integer.

Sample Input 1

5 7 4

Sample Output 1

2

When x=3, floor(Ax/B)-A×floor(x/B) = floor(15/7) - 5×floor(3/7) = 2. This is the maximum value possible.

Sample Input 2

11 10 9

Sample Output 2

9

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;

int main(){
  int64_t A, B, N, ans;
  cin >>A >> B >> N;
  ans = 0;
  if(N > 1000000 && A > 100000)
  {
  for(int64_t i = 1; i <= N -A/B ;i++){
    int64_t x = A*i/B -A*(i/B);
    if(ans < x){
      ans = x;
    }
  }
  }
  else
  {
  for(int64_t i = N-2*A/B-1; i <= N;i++){
    int64_t x = A*i/B -A*(i/B);
    if(ans < x){
      ans = x;
    }
  }
  }
  cout << ans << endl;

}