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;

/**
 * Created by aaabdulh on 09/05/17.
 */
public class Main {
    private static final int MAX = 1000 * 1000 * 1000;

    public static void main(String[] args) {
        int a, b, p, q, n;
        Scanner input = new Scanner(System.in);
        n = input.nextInt();
        while (n-- > 0) {
            a = input.nextInt();
            b = input.nextInt();
            p = input.nextInt();
            q = input.nextInt();

            long denum = b + q - b % q;
            long minDenum = denum;
            long l, r, mid = 0;
            r = MAX;
            l = 1;

            while (l < r) {
                mid = (l + r) / 2;
                denum = q * mid;
                long num = p * mid;
                long denumDiff = denum - b;
                long numDiff = num - a;
                if (denumDiff >= numDiff && numDiff >= 0) {
                    r = mid;
                } else {
                    l = mid + 1;
                }
            }
            long bb = q*l;
            long aa = p*l;
//            System.out.println(aa + " " + bb);
            System.out.println(bb - b >= aa - a && aa - a >= 0 ?bb-b:-1);
        }
    }
}
