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.
//package CF_412;

import java.io.*;
import java.util.*;

public class C {
	public static void main(String[] args){
		Scanner sc=new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int t=sc.nextInt();
		for(int a=0; a<t; a++){
			int x=sc.nextInt(), y=sc.nextInt(), p=sc.nextInt(), q=sc.nextInt();
			if(p==0)
				System.out.println(x==0?0+"":-1+"");
			else if(p==q)
				System.out.println(x==y?0+"":-1+"");
			else{
				int t1=(x+p-1)/p;
				int t2=((y-x)+((q-p)-1))/(q-p);
				System.out.println(q*1L*Math.max(t1, t2)-y);
			}
		}
	}
}
