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.lang.*;
import java.util.*;

public final class Solution {
	public static void main(String[] args) {
		Solution p = new Solution();
		p.run();
	}
	
	private void run() {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		
		for (int i = 0; i< t; i++) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			int p = sc.nextInt();
			int q = sc.nextInt();
			
		    System.out.println(solve(x,y,p,q));
		}
	}
	
	private long solve(int x, int y, int p, int q) {
		if (p == 0) {
			if (x != 0) {
				return -1;
			} else {
				return 0;
			}
		}
		if (p == q) {
			if (x != y) {
				return -1;
			} else {
				return 0;
			}
		}
		
		long k = (long)Math.ceil(Math.max(x*1.0/p, ((y-x)*1.0/(q-p))));

		return ((k*q) - y);
	}
}