#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    long long l = 1, r = INT_MAX;
    long long ans = -1;
    function<long long(long long)> check = [&](long long k) {
      long long b = q * k - y;
      long long a = p * k - x;
      if (((a + x) * q) == ((b + y) * p) && a >= 0 && b >= 0 && b >= a)
        return b;
      return (long long)-1;
    };
    while (l <= r) {
      long long mid = (l + r) / 2;
      long long res = check(mid);
      if (res != -1) {
        r = mid - 1;
        ans = res;
      } else
        l = mid + 1;
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    long long l = 1, r = INT_MAX;
    long long ans = -1;
    function<long long(long long)> check = [&](long long k) {
      long long b = q * k - y;
      long long a = p * k - x;
      if (((a + x) * q) == ((b + y) * p) && a >= 0 && b >= 0 && b >= a)
        return b;
      return (long long)-1;
    };
    while (l <= r) {
      long long mid = (l + r) / 2;
      long long res = check(mid);
      if (res != -1) {
        r = mid - 1;
        ans = res;
      } else
        l = mid + 1;
    }
    cout << ans << "\n";
  }
  return 0;
}

```