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.
#include <bits/stdc++.h>
using namespace std;
long long x, y, p, q;
int main() {
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> x >> y >> p >> q;
    if ((p / q == 1 && x < y) || (p == 0 && x > 0))
      cout << -1 << "\n";
    else {
      if (x * q == p * y)
        cout << 0 << "\n";
      else {
        long long l = 0, z = 1000000001;
        while (l + 1 < z) {
          long long s = (l + z) / 2;
          long long act = s * p;
          long long bct = s * q;
          if (bct - y < act - x || act < x) {
            l = s;
          } else {
            z = s;
          }
        }
        cout << z * q - y << "\n";
      }
    }
  }
  return 0;
}
