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.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Div2_412C {

	public static void main(String[] args) throws IOException {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter printer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int numT = Integer.parseInt(reader.readLine());

		for (int i = 0; i < numT; i++) {
			StringTokenizer inputData = new StringTokenizer(reader.readLine());
			long x = Long.parseLong(inputData.nextToken());
			long y = Long.parseLong(inputData.nextToken());
			long p = Long.parseLong(inputData.nextToken());
			long q = Long.parseLong(inputData.nextToken());

			if (q == p) {
				printer.println(x == y ? 0 : -1);
				continue;
			}

			if (p == 0) {
				printer.println(x == 0 ? 0 : -1);
				continue;
			}

			long b1 = x == 0 ? 0 : (x - 1) / p + 1;
			long b2 = y == 0 ? 0 : (y - 1) / q + 1;
			long b3 = x == y ? 0 : (y - x - 1) / (q - p) + 1;
			long m = Math.max(b1, Math.max(b2, b3));
			printer.println(m * q - y);
		}
		printer.close();
	}

}
