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: Runtime Error)

import java.math.BigDecimal;
import java.math.RoundingMode;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		BigDecimal A = new BigDecimal(sc.next());
		BigDecimal B = new BigDecimal(sc.next());
		BigDecimal N = new BigDecimal(sc.next());
		BigDecimal n = BigDecimal.ZERO;
		BigDecimal E = N.multiply(A).divide(B);
		BigDecimal MAX = BigDecimal.ZERO;
		while (n.compareTo(E) != 1) {
			BigDecimal R = n.subtract(n.divide(A, 0, RoundingMode.DOWN).multiply(A));
			if (R.compareTo(MAX) == 1) {
				MAX = R;
			}
			n = n.add(BigDecimal.ONE);
		}
		System.out.println(MAX);
	}

}