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.math.BigInteger;
import java.util.Scanner;

/**
 * Created by yukesh on 5/24/2017.
 */
public class p807c {
    public static void main(String[] args) {
        double p,q,x,y,k,a;
        BigInteger s;
        int t;
        Scanner sc=new Scanner(System.in);
        t=sc.nextInt();
        for (int i = 0; i <t; i++) {
        x=(double)sc.nextInt();
        y=(double)sc.nextInt();
        p=(double)sc.nextInt();
        q=(double)sc.nextInt();
        if(q==p&&x==y) {
            System.out.println(0);
            continue;
        }
            if(p==0&&x==0)
            {
                System.out.println(0);
                continue;
            }
            if(q==p&&x!=y) {
                System.out.println(-1);
                continue;
            }
            if(p==0&&x!=0) {
                System.out.println(-1);
                continue;
            }
        k=Math.ceil((y-x)/(q-p));
        if(k<=Math.ceil(x/p))
        {
            a=Math.ceil(x/p)*p-x;
            s=BigInteger.valueOf((long)q).multiply(BigInteger.valueOf((long)Math.ceil(x/p))).subtract(BigInteger.valueOf((long)y));


        }
        else
        {
            a=k*p-x;
            s= BigInteger.valueOf((long) q).multiply(BigInteger.valueOf((long)k)).subtract(BigInteger.valueOf((long)y));

        }
        System.out.println(s);
        }


    }
}
