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 scan = new Scanner(System.in);
        long a = scan.nextLong();
        long b = scan.nextLong();
        long n = scan.nextLong();
		List<Long> listA = new ArrayList<Long>(); 


		if(calculate(a, b, 1) > calculate(a,b,n)){
			for (long x=1; x<(n+1)/2; x++) {
				listA.add(calculate (a, b, x));
	        }
        } else {
			for (long x=0; x<(n+1)/2; x++) {
				listA.add(calculate (a, b, (n-x)));
	        }
        }

		Collections.sort(listA,  Collections.reverseOrder());
		System.out.println(listA.get(0));
	    scan.close();	
    }
    
    public static long calculate(long a, long b, long x) {
    	return 	(long)((a*x)/b) - a*(long)(x/b);
	}
}