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 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long A = Long.parseLong(sc.next());
		long B = Long.parseLong(sc.next());
		long N = Long.parseLong(sc.next());
		long max = -1;
		long maxX = N;
		long minX = 0;
		long bb = 1;
		while(true) {
			long a = (A * bb /B) - A * (bb/B);
			if(a > max) {
				max = a;
				minX = bb;
			} else if(a == max) {
				break;
			} else {
				maxX = bb;
			}
			bb = (maxX + minX )/2;
		}
		for(long i=bb;i<=N;i++) {
			long a = (A * i /B) - A * (i/B);
			if(a >= max) {
				max = a;
			} else {
				break;
			}
//			max = Math.max(max, a);
		}
		System.out.println(max);
	}
}