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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main{

	public static void main(String[] args) {

		Scanner scanner = new Scanner(System.in);

		String[] tmp = scanner.nextLine().split(" ");
		long a = Long.parseLong(tmp[0]);
		long b = Long.parseLong(tmp[1]);
		long n = Long.parseLong(tmp[2]);


		long minmax = Math.min(n, b - 1);
		if(minmax < b) {

		long zero = (long)(Math.floor((a * minmax) / b) + a * Math.floor(minmax / b));

		long max  = (long)(Math.floor((a * n) / b) + a * Math.floor(n/ b));

		System.out.println(Math.max(max, zero));
		return;
		}

		long max  = (long)(Math.floor((a * n) / b) + a * Math.floor(n/ b));

		System.out.println(max);




	}

}