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.*;
import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		
		int a = scan.nextInt();
		long b = scan.nextLong();
		long n = scan.nextLong();
		
		if(n<b) {
			double x = Math.floor(a*n/b)-5* Math.floor(n/b);
			System.out.println((int)x);
			
		}else {
			long y = b-1;
			double x = Math.floor(a*y/b)-5* Math.floor(y/b);
			System.out.println((int)x);
		}
		
	}	
	
	
	//最小公約数
	/*
	static int gcd (int a, int b) {
		int temp;
		while((temp = a%b)!=0) {
			a = b;
			b = temp;
		}
		return b;
	}
	*/
	
}