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 C {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0){
			x = sc.nextInt(); y = sc.nextInt();
			p = sc.nextInt(); q = sc.nextInt();
			
			if(p == 0 && x != 0 || p == q && x != y) {
				System.out.println(-1);
				continue;
			}

			int low = 0;
			int high = (int) 1e9;
			long ans = -1;
			while(low <= high) {
				int mid = (low + high) / 2;
				long den = (long) mid * q;
				long num = (long) mid * p;
				
				if(den >= y && num >= x && den - y >= num - x) {
					
					ans = den - y;
					high = mid - 1;
				}
				else
					low = mid + 1;
			}
			
			System.out.println(ans);
		}

	}
	static int x, y, p, q;
	static boolean can(int sub) {
		int nu = p;
		int du = q;

		while(nu + du <= x + y + sub) {
			if(x + y + sub == nu + du)
				return true;
			nu += p;
			du += q;
		}

		return false;
	}
}
