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.
/*
Constraints:
y >= y0
y = k * q, k = 0, 1, 2, ...
x = k * p
x >= x0
x <= x0 + y - y0

==>
k >= x0 / p
k >= (y0-x0) / (q-p)
k >= y0 / q
*/

import java.util.Scanner;

public class P807CSuccessRate {
    private long div_ceil(long a, long b) {
        return (int) Math.ceil(((double) a) / b);
    }

    long calc(long x, long y, long p, long q) {
        if (q == 0) {
            if (x == 0 && y == 0 && p == 0) {
                return 0;
            } else {
                return -1;
            }
        }
        long k3 = div_ceil(y, q);

        long k1;
        if (p == 0) {
            if (x != 0) {
                return -1;
            } else {
                k1 = Integer.MIN_VALUE;
            }
        } else {
            k1 = div_ceil(x, p);
        }

        long k2;
        if (p == q) {
            if (x != y) {
                return -1;
            } else {
                k2 = Integer.MIN_VALUE;
            }
        } else {
            k2 = div_ceil(y - x, q - p);
        }

        long k = Math.max(Math.max(k1, k2), k3);
        return k * q - y;
    }

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();

        P807CSuccessRate obj = new P807CSuccessRate();
        for (int i = 0; i < n; ++i) {
            int x = scanner.nextInt();
            int y = scanner.nextInt();
            int p = scanner.nextInt();
            int q = scanner.nextInt();
            long res = obj.calc(x, y, p, q);
            System.out.println(res);
        }
    }
}
