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 C807 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		for (int i = 0; i < n; i++) {
			long x = sc.nextLong();
			long y = sc.nextLong();
			long p = sc.nextLong();
			long q = sc.nextLong();
			
			if (p==0) {
				if (x==0) {
					System.out.println("0");
				} else {
					System.out.println("-1");
				}
				continue;
			}
			
			if (p==q) {
				if (x==y) {
					System.out.println("0");
				} else {
					System.out.println("-1");
				}
				continue;
			}

			long base_mult = y/q==0?1:y/q;
			long mult;
			do {
				for (mult = 1; ; mult=mult*2) {
					long diffT = (base_mult+mult)*q-y;
					long diffW = (base_mult+mult)*p-x;
					if (diffW >= 0 && diffW <= diffT) {
						//System.out.println(diffT);
						break;
					}
				}
				
				base_mult += mult/2;
			} while (mult != 1);
			
			for (; ; base_mult++) {
				long diffT = base_mult*q-y;
				long diffW = base_mult*p-x;
				if (diffW >= 0 && diffW <= diffT) {
					System.out.println(diffT);
					break;
				}
			}
		}
	}
}
