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: Accepted)

import java.util.Scanner;

public class Main {
	public static void main(String[]args) {
		try(Scanner scan = new Scanner(System.in)){

			long A = scan.nextLong();
			long B = scan.nextLong();
			long N = scan.nextLong();

			/*
			for(int x = 0;x<=N;x++) {
				long t = (A*x)/B;
				long s = x/B;

				System.out.println(t-A*s);
			}*/

			long t,s;

			if(N<B-1) {
				 t = (A*N)/B;
				 s = N/B;

			}else {
				 t = (A*(B-1))/B;
				 s = (B-1)/B;
			}


			System.out.println(t-A*s);

		}


	}


}