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.*;
import java.util.*;
public class Main {
	static double eps=1e-12;
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter pw = new PrintWriter(System.out);
		int n=sc.nextInt();
		while (sc.hasNext()) {
			long x=sc.nextInt();
			long y=sc.nextInt();
			long p=sc.nextInt();
			long q=sc.nextInt();
			if(p==0){
				if(x==0){
					pw.println(0);
				}else{
					pw.println(-1);
				}
			}
			else if(p==q){
				if(x==y){
					pw.println(0);
				}else{
					pw.println(-1);
				}
			}else{
				long start=0;
				long end=(long) 1e9;
				long result=end;
				while(start<=end){
					long mid=(start+end)/2;
					if(check(x,y,p*mid,q*mid)){
						result=mid;
						end=mid-1;
					}else{
						start=mid+1;
					}
				}
				if(result==end){
					pw.println(-1);
				}else
					pw.println(Math.abs(y-q*result));
			}
			pw.flush();
		}
	}
	static boolean check(long x,long y,long p,long q){
		if(q-y>=0&&p-x>=0&&q-y>=p-x)return true;
		else return false;
	}
}

class FastScanner {
	BufferedReader br;
	StringTokenizer st;

	public FastScanner() {
		try {
			br = new BufferedReader(new InputStreamReader(System.in));
			st = new StringTokenizer("");
		} catch (Exception e) {
			e.printStackTrace();
		}
	}

	public boolean hasNext() {
		while (!st.hasMoreTokens()) {
			String line = nextLine();
			if (line == null) {
				return false;
			}
			st = new StringTokenizer(line);
		}
		return true;
	}

	public String next() {
		if (st.hasMoreTokens())
			return st.nextToken();
		try {
			st = new StringTokenizer(br.readLine());
		} catch (Exception e) {
			e.printStackTrace();
		}
		return st.nextToken();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public String nextLine() {
		String line = "";
		try {
			line = br.readLine();
		} catch (Exception e) {
			e.printStackTrace();
		}
		return line;
	}
}