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: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) {
		FastScanner sc = new FastScanner(System.in);
		long a = sc.nextLong(), b = sc.nextLong(), n = sc.nextLong();
		long ans = 0l;
		int c = 0;
		long k = n;
		if(n > a) k = a;
		for(long i = k; i > 0; i--) {
			long x = (a*i/b) - a*(i/b);
			if(x > ans) {
				ans = x;
				c++;
			}else if(c != 0 && x <= ans) {
				break;
			}
		}
		if(c == 1) {
			for(long i = k+1; i <= n; i++) {
				long x = (a*i/b) - a*(i/b);
				if(x > ans) {
					ans = x;
					c++;
				}else if(c != 0 && x < ans) {
					break;
				}
			}
		}
		System.out.println(ans);
	}
	
	static class FastScanner {
		private BufferedReader reader = null;
	    private StringTokenizer tokenizer = null;
	    public FastScanner(InputStream in) {
	        reader = new BufferedReader(new InputStreamReader(in));
	        tokenizer = null;
	    }

	    public String next() {
	        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                tokenizer = new StringTokenizer(reader.readLine());
	            } catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }
	        return tokenizer.nextToken();
	    }

	    public String nextLine() {
	        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                return reader.readLine();
	            } catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }

	        return tokenizer.nextToken("\n");
	    }

	    public long nextLong() {
	        return Long.parseLong(next());
	    }

	    public int nextInt() {
	        return Integer.parseInt(next());
	    }

	    public double nextDouble() {
	         return Double.parseDouble(next());
	     }

	    public int[] nextIntArray(int n) {
	        int[] a = new int[n];
	        for (int i = 0; i < n; i++)
	            a[i] = nextInt();
	        return a;
	    }

	    public long[] nextLongArray(int n) {
	        long[] a = new long[n];
	        for (int i = 0; i < n; i++)
	            a[i] = nextLong();
	        return a;
	    }
	}
}