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.lang.*;
import java.io.*;
public class Main
{
    static boolean check(long x,long y,long p,long q,long t){
        if(p*t>=x&&(q-p)*t>=(y-x))
        return true;
        else 
        return false;
        
    }
	public static void main (String[] args) throws java.lang.Exception
	{
	    Scanner sc=new Scanner(System.in);
	   // BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    int t=1;
	    t=sc.nextInt();
	  //  int t=Integer.parseInt(br.readLine());
	    while(--t>=0){
	        long x=sc.nextLong();
	        long y=sc.nextLong();
	        long p=sc.nextLong();
	        long q=sc.nextLong();
	        long l=0,r=1000000000;
	        if(!check(x,y,p,q,r)){
	            System.out.println(-1);
	            continue;
	        }
	        long min=1000000000;
	       while(l<=r){
	           long m=(l+r)/2;
	           if(check(x,y,p,q,m)){
	               min=Math.min(min,m);
	               r=m-1;
	           }
	           else{
	               l=m+1;
	           }
	       }
	       System.out.println(q*min-y);
	        
	    }
	    
	}
}
