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.Scanner;

public class Main {
    public static boolean check(long x, long y, long p, long q, long t) {
        return p * t >= x && q * t - p * t >= y - x;
    }
    public static void main(String args[]){
        Scanner scanner=new Scanner(System.in);
       long t=scanner.nextInt();
        for(int k=0;k<t;k++){
           long x,y,p,q;
            x=scanner.nextInt();
            y=scanner.nextInt();
            p=scanner.nextInt();
            q=scanner.nextInt();

            long r=(int)Math.pow(10,9);
            long l=0;
            if (!check(x, y, p, q, r)) {
                System.out.println(-1);
                continue;
            }

            while (r-l>1){
               long m=l+(r-l)/2;
               if(check(x,y,p,q,m)){
                   r=m;
               }
               else {
                   l=m;
               }
            }
            System.out.println(r*q-y);

        }
    }
}