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.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.io.PrintWriter;
import java.io.File;

public class Solution {
    public static void main(String[] args) throws Exception {
        Solution solution = new Solution();
        solution.run();
    }

    private void run() throws Exception {
        Scanner scan = new Scanner(System.in);
        PrintWriter writer = new PrintWriter(System.out);
        int n = scan.nextInt();


        for (int i = 0; i < n; i++) {
            long x = scan.nextLong();
            long y = scan.nextLong();
            long p = scan.nextLong();
            long q = scan.nextLong();
            if (p == q && x != y || p == 0 && x!= 0 ) {
                writer.println(-1);
            } else {

                long tp = p;
                long tq = q;
                long M = Math.max(x*2,y*2);
                long l = 0;
                long r = M;
                int k = 2;
                long ansp = 0;
                long ansq = 0;
                boolean fl = true;
                while (l <= r&& fl) {
                    long med = l + (r - l) / 2;
                    ansp = p * med - x;
                    ansq = q * med - y;
                    if (ansp >= 0 && ansq >= 0 && ansq >= ansp) {
                        r = med;
                    } else {
                        l = med;
                    }
                    if (r - l < 2) {
                        for (long f = l; f <= r; f++) {
                            ansp = p * f - x;
                            ansq = q * f - y;
                            if (ansp >= 0 && ansq >= 0 && ansq >= ansp) {
                               fl = false;
                            }
                        }
                    }
                }

                long ans =  ansq;
                writer.println(ans);
            }


        }


        writer.close();
    }


}