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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class C {
	public static void main(String[] args) throws IOException{
		FastScanner in = new FastScanner(System.in);
		PrintWriter out= new PrintWriter(System.out);
		int n= in.nextInt();
		for (int i = 0; i < n; i++) {
			long a= in.nextLong();
			long b= in.nextLong();
			long p= in.nextLong();
			long q= in.nextLong();
			if(p==0&&a==0){
				System.out.println(0);
				continue;
			}
			if(p==q&&a==b){
				System.out.println(0);
				continue;
			}
			if(p==q||(p==0&&a!=0)){
				System.out.println(-1);
				continue;
			}
			int res= (int) (b/q);
			if(b%q!=0) res++;
			int res1= (int) (a/p);
			if(a%p!=0) res1++;
			res= Math.max(res, res1);
			int res2= (int) ((b-a)/(q-p));
			if((b-a)%(q-p)!=0) res2++;
			res=  Math.max(res,res2);
			System.out.println(res*q-b);
			
			//binary search j
			
			
		}
	}
	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
			st = new StringTokenizer("");
		}

		public String next() throws IOException {
			if (!st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
				return next();
			}
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
	}
}