#include <bits/stdc++.h>
using namespace std;
long long x, y, p, q;
long long const inf = 1e9;
long long success_rate() {
  long long ans = (long long)-1;
  long long left = 0;
  long long right = (long long)inf;
  while (left <= right) {
    long long k = (long long)(left + right) / 2;
    long long p1, q1;
    p1 = k * p;
    q1 = k * q;
    long long x1, y1;
    x1 = p1 - x;
    y1 = q1 - y;
    if (x1 >= 0 && x1 <= y1) {
      ans = y1;
      right = k - 1;
    } else {
      left = k + 1;
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    cout << success_rate() << endl;
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long x, y, p, q;
long long const inf = 1e9;
long long success_rate() {
  long long ans = (long long)-1;
  long long left = 0;
  long long right = (long long)inf;
  while (left <= right) {
    long long k = (long long)(left + right) / 2;
    long long p1, q1;
    p1 = k * p;
    q1 = k * q;
    long long x1, y1;
    x1 = p1 - x;
    y1 = q1 - y;
    if (x1 >= 0 && x1 <= y1) {
      ans = y1;
      right = k - 1;
    } else {
      left = k + 1;
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    cout << success_rate() << endl;
  }
  return 0;
}

```