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.
/**
 * Codeforces, Problem set, 588 B
 * 
 * Caleb Ji 
 * Finds largest squarefree divisor
 */

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

public class constant{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        
        
        for(int i = 0; i < n; i++){
            long x = sc.nextLong();
        long y = sc.nextLong();
        long p = sc.nextLong();
        long q = sc.nextLong();
            System.out.println(blah(x, y, p, q));
        }

        
    }
    
    public static long blah(long x, long y, long p, long q){
        //System.out.println(q + " " + x + " " + p + " " + y);
        if(q*x==p*y) return 0;
        if(p == 0 || p == q) return -1;
        long num =1+ Math.max(Math.max((y-1)/q, (x-1)/p), (y-x-1)/(q-p));
        //System.out.println((y-1)/q);
        //inSystem.out.println(num + " " + (y-x-1)/(q-p));
        return num*q-y;
    } 
        
}