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 Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {

  public static void main(String[] args) { 
      Scanner sc = new Scanner(System.in);
      long a = sc.nextInt();
      long b = sc.nextLong();
      long n = sc.nextLong();
      long left = 0, right = n;
      while(left <= right) {
        long mid1 = left + (right - left) / 3;
        long mid2 = left + 2 * (right - left) / 3;
        if(cal(mid1, a, b) <= cal(mid2, a, b)) {
          left = mid1 + 1;
        } else right = mid2 - 1;
      }
      System.out.println(cal(right < 0 ? 0 : right, a, b));
  }

  private static long cal(long x, long a, long b) {
    return a * x / b - a * (x / b);
  }
}