You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made y submissions, out of which x have been successful. Thus, your current success rate on Codeforces is equal to x / y.

Your favorite rational number in the [0;1] range is p / q. Now you wonder: what is the smallest number of submissions you have to make if you want your success rate to be p / q?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each of the next t lines contains four integers x, y, p and q (0 ≤ x ≤ y ≤ 109; 0 ≤ p ≤ q ≤ 109; y > 0; q > 0).

It is guaranteed that p / q is an irreducible fraction.

Hacks. For hacks, an additional constraint of t ≤ 5 must be met.

Output

For each test case, output a single integer equal to the smallest number of submissions you have to make if you want your success rate to be equal to your favorite rational number, or -1 if this is impossible to achieve.

Example

Input

4
3 10 1 2
7 14 3 8
20 70 2 7
5 6 1 1


Output

4
10
0
-1

Note

In the first example, you have to make 4 successful submissions. Your success rate will be equal to 7 / 14, or 1 / 2.

In the second example, you have to make 2 successful and 8 unsuccessful submissions. Your success rate will be equal to 9 / 24, or 3 / 8.

In the third example, there is no need to make any new submissions. Your success rate is already equal to 20 / 70, or 2 / 7.

In the fourth example, the only unsuccessful submission breaks your hopes of having the success rate equal to 1.
import java.util.*;
import java.io.*;

public class SuccessRate 
{
	static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(new
                     InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt()
        {
            return Integer.parseInt(next());
        }
 
        long nextLong()
        {
            return Long.parseLong(next());
        }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
	
	static long x, y, p, q, limit = Long.parseLong("3000000000");
	static long e19 = 1000000000;
	
	public static void main(String[] args) 
	{
		FastReader reader = new FastReader();
		PrintWriter writer = new PrintWriter (System.out);
		int testCases = reader.nextInt();
		
		while (testCases > 0)
		{
			x = reader.nextInt();
			y = reader.nextInt();
			p = reader.nextInt();
			q = reader.nextInt();
			
			if (!check(e19))
				writer.println (-1);
			else	
				writer.println (binarySearch(0, limit));
		
			testCases--;
		}
		
		writer.close();
	}
	
	static long binarySearch (long low, long high)
	{
		while (low < high)
		{
			long mid = (low+high)/2;
	
			//long a = mid*p - x;
			if (!check(mid))
				low = mid+1;
			else
				high = mid;
		}
		
		return (low*q - y);
	}

	static boolean check (long t)
	{
		boolean check1 = (p*t >= x);
		boolean check2 = (q*t - p*t >= y-x);
		
		return (check1 && check2);
	}
}
