#include <bits/stdc++.h>
using namespace std;
long long x, y, p, q, st, dr;
long long posibil(long long ans) {
  long long a, b;
  a = p * ans - x;
  b = q * ans - a - y;
  if (a >= 0 && b >= 0) return a + b;
  return -1;
}
void solve() {
  cin >> x >> y >> p >> q;
  if (p == q && x != y) {
    cout << "-1\n";
    return;
  }
  st = 1;
  dr = 1e9;
  while (st < dr) {
    long long mid = (st + dr) / 2;
    if (posibil(mid) != -1)
      dr = mid;
    else
      st = mid + 1;
  }
  if (posibil(st) >= 0 || posibil(dr) >= 0) {
    long long min1, min2;
    min1 = posibil(st);
    min2 = posibil(dr);
    if (min1 < 0)
      cout << min2 << '\n';
    else if (min2 < 0)
      cout << min1 << '\n';
    else
      cout << min(min1, min2) << '\n';
  } else
    cout << "-1\n";
}
int main() {
  long long t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Generate a CPP solution 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, st, dr;
long long posibil(long long ans) {
  long long a, b;
  a = p * ans - x;
  b = q * ans - a - y;
  if (a >= 0 && b >= 0) return a + b;
  return -1;
}
void solve() {
  cin >> x >> y >> p >> q;
  if (p == q && x != y) {
    cout << "-1\n";
    return;
  }
  st = 1;
  dr = 1e9;
  while (st < dr) {
    long long mid = (st + dr) / 2;
    if (posibil(mid) != -1)
      dr = mid;
    else
      st = mid + 1;
  }
  if (posibil(st) >= 0 || posibil(dr) >= 0) {
    long long min1, min2;
    min1 = posibil(st);
    min2 = posibil(dr);
    if (min1 < 0)
      cout << min2 << '\n';
    else if (min2 < 0)
      cout << min1 << '\n';
    else
      cout << min(min1, min2) << '\n';
  } else
    cout << "-1\n";
}
int main() {
  long long t;
  cin >> t;
  while (t--) solve();
}

```