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: Time Limit Exceeded)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long a = sc.nextLong();
		double b = sc.nextDouble();
		long c = sc.nextLong();
		long max = Long.MIN_VALUE;
		long max2 = Long.MIN_VALUE;
		for (long i = c; i >= 0; --i) {
			long tmp = (long) (a * i / b) - (long) (a * (long) (i / b));
			if (tmp >= max)  {
				max = tmp;
			} else {
				break;
			}
		}
		for (long i = 0; i <= c; ++i) {
			long tmp = (long) (a * i / b) - (long) (a * (long) (i / b));
			if (tmp >= max2)  {
				max2 = tmp;
			} else {
				break;
			}
		}
		System.out.println(Math.max(max, max2));
	}
}