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 TestClass {

	static PrintWriter out = new PrintWriter(System.out);
	public static void main(String[] args) throws Exception{
		
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(in.readLine());
		for(int i=0;i<n;i++)
		{
			String s[] = in.readLine().split(" ");
			long x = Long.parseLong(s[0]);
			long y = Long.parseLong(s[1]);
			long p = Long.parseLong(s[2]);
			long q = Long.parseLong(s[3]);
			if((p==q && x!=y) || (p==0 && x!=0))
			{
				out.println(-1);
				continue;
			}
			long l=0,h=(long)1e9;
			while(l<=h)
			{
				long m = l + (h-l)/2;
				if(p*m-x>=0 && q*m+x-p*m-y>=0)
				{
					h = m-1;
				}
				else
				{
					l = m+1;
				}
			}
			out.println(q*l-y);
		}
		out.close();
	}
	
}	