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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public final class Main {

    public static void main(String[] args) {
        final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        final long a = in.nextLong();
        final long b = in.nextLong();
        final long n = in.nextLong();
        in.nextLine();
        long lo = 1;
        long hi = n;
        while (lo < hi) {
            long mid = lo + hi >>> 1;
            long f = f(a, mid, b);
            long f1 = f(a, mid - 1, b);
            if (f >= f1) {
                lo = mid + 1;
            } else {
                hi = mid;
            }
        }
        System.out.println(f(a, lo, b));
    }

    private static long f(long a, long mid, long b) {
        return Math.floorDiv(a * mid, b) - a * Math.floorDiv(mid, b);
    }
}