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 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int a = sc.nextInt();
        long b = sc.nextLong();
        long n = sc.nextLong();

        if (b == 1) {
            System.out.println(0);
        } else {
            if (b > n) {
                System.out.println((long) ((double) a / (double) b * n) - (long) ((double) 1 / (double) b * n) * a);
            } else if (b == n) {
                System.out.println(
                        ((long) ((double) a / (double) b * (n - 1))) - (long) ((double) 1 / (double) b * (n - 1)) * a);
            } else { /* b<n */
                if (n % b == 0) {
                    System.out.println(((long) ((double) a / (double) b * (n - 1)))
                            - (long) ((double) 1 / (double) b * (n - 1)) * a);
                } else {
                    System.out.println((long) ((double) a / (double) b * n) - (long) ((double) 1 / (double) b * n) * a);
                }
            }
        }
        sc.close();
    }
}