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.awt.Point;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Solution {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String in = br.readLine();
		StringTokenizer st = new StringTokenizer(in);
		int t = Integer.parseInt(st.nextToken());
		for (int i = 0; i < t; ++ i) {
			st = new StringTokenizer(br.readLine());
			long x, y, p, q;
			x = Long.parseLong(st.nextToken());
			y = Long.parseLong(st.nextToken());
			p = Long.parseLong(st.nextToken());
			q = Long.parseLong(st.nextToken());
			
			long start = -1, end = (long) 1e9;
			
			if (! (end * p >= x && end * q - end * p >= y - x)) {
				System.out.println(-1);
			} else {
				while (end - start > 1) {
					long c = (start + end) / 2;
					if (c * p >= x && c * q - c * p >= y - x) {
						end = c;
					} else {
						start = c;
					}
				}
				System.out.println(q * end - y);
			}
		}
	}
}