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.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
	static long x, y, p, q;

	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;
		StringBuilder sb = new StringBuilder();
		int t = Integer.parseInt(br.readLine());
		while (t-- > 0) {
			st = new StringTokenizer(br.readLine());
			x = Long.parseLong(st.nextToken());
			y = Long.parseLong(st.nextToken());
			p = Long.parseLong(st.nextToken());
			q = Long.parseLong(st.nextToken());
			if (p == q)
				if (x == y)
					sb.append(0).append('\n');
				else
					sb.append(-1).append('\n');
			
			else if (p == 0)
				if (x == 0)
					sb.append(0).append('\n');
				else
					sb.append(-1).append('\n');

			else {
				long f = binarySearch(1L, 1000000000L);
				sb.append(f * q - y).append('\n');
			}

		}
		System.out.print(sb);
	}

	private static long binarySearch(long l, long r) {
		if (l == r)
			return l;

		long mid = (l + r) / 2;
		if (check(mid))
			return binarySearch(l, mid);
		else
			return binarySearch(mid + 1, r);
	}

	private static boolean check(long f) {
		return x <= p * f && y - x <= (q - p) * f;
	}
}
