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 P807C
{
	public static void main(String[] args)
	{
		Scanner scan = new Scanner(System.in);
		int t = scan.nextInt();
		for (int z = 0; z < t; z++)
		{
			int x = scan.nextInt();
			int y = scan.nextInt();
			int p = scan.nextInt();
			int q = scan.nextInt();
			if (p == 0)
			{
				System.out.println(x == 0 ? 0 : -1);
				continue;
			}
			if (p == 1 && q == 1)
			{
				System.out.println(x == y ? 0 : -1);
				continue;
			}
			long lo = 0, hi = (int)1e9;
			long gx = 0, gy = 0;
			while (lo < hi)
			{
				long mid = (lo+hi)/2;
				gx = p*mid;
				gy = q*mid;
				if (gy < y || gx < x)
				{
					lo = mid+1;
					continue;
				}
				long dx = gx-x;
				long dy = gy-y;
				if (dx <= dy)
					hi = mid;
				else
					lo = mid+1;
			}
			System.out.println(q*lo - y);
		}
	}
}
