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: Time Limit Exceeded)

import java.util.Scanner;
class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		/*
		int a = sc.nextInt();
		int b = sc.nextInt();
		int c = sc.nextInt();
		int[][] d = new int[c][4];
		int[] e = new int[a];
		for(int i = 0; i < c; i++) {
			for(int j = 0; j < 4; j++) {
				d[i][j] = sc.nextInt();
			}
		}
		for(int i = 0; )
		*/
		int a = sc.nextInt();
		long b = sc.nextLong();
		long c = sc.nextLong();
		long d = 0;
		long e;
		int f = 0;
		for(int i = 1; i <= c; i++) {
			e =(a*i)/b-a*(i/b);
			if(e>d) {
				d = e;
			}
		}
		System.out.println(d);
	}
}