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.*;

class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int A = Integer.parseInt(sc.next());
        long B = Long.parseLong(sc.next());
        long N = Long.parseLong(sc.next());
        long amari = N % B;
        if(N < B) {
            System.out.println(floor(A*N/B) - (A * floor(N/B)));
        }else {
            if(floor(A/B) - A * floor(1/B) >= floor(A) - A * floor(1)) {
                System.out.println(floor((A * (B-1))/B) - A * floor((B-1)/B));
            }else if(floor(A/B) - A * floor(1/B) < floor(A) - A * floor(1)) {
                if(floor((A * (N - amari - 1))/B) - A * floor((N - amari - 1)/B) < floor((A * N) / B) - A * floor(N/B)) {
                    System.out.println(floor((A * N) / B) - A * floor(N/B));
                }else {
                    System.out.println(floor((A * (N - amari - 1))/B) - A * floor((N - amari - 1)/B));
                }
            }
        }
        
    }
    
    public static long floor(double t) {
        return (long)t;
    }
    
    public static long floor(long t) {
        return (long)t;
    }
}