#include <bits/stdc++.h>
using namespace std;
long long a[1001];
bool moguce(int x, int y, int p, int q, int s) {
  long long p1 = (long long)p * s, p2 = (long long)q * s;
  if (p1 >= (long long)x && p2 >= (long long)y &&
      p1 - (long long)x <= (long long)p2 - y)
    return true;
  else
    return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, x, y, p, q, i, poc, kraj, s;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x >> y >> p >> q;
    if (moguce(x, y, p, q, 1000000001) == false)
      a[i] = -1;
    else {
      poc = 1;
      kraj = 1000000001;
      while (poc <= kraj) {
        s = (poc + kraj) / 2;
        if (moguce(x, y, p, q, s) == true && moguce(x, y, p, q, s - 1) == false)
          break;
        else {
          if (moguce(x, y, p, q, s) == false)
            poc = s + 1;
          else
            kraj = s - 1;
        }
      }
      a[i] = (long long)q * s - y;
    }
  }
  for (i = 0; i < n; i++) cout << a[i] << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 a[1001];
bool moguce(int x, int y, int p, int q, int s) {
  long long p1 = (long long)p * s, p2 = (long long)q * s;
  if (p1 >= (long long)x && p2 >= (long long)y &&
      p1 - (long long)x <= (long long)p2 - y)
    return true;
  else
    return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, x, y, p, q, i, poc, kraj, s;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x >> y >> p >> q;
    if (moguce(x, y, p, q, 1000000001) == false)
      a[i] = -1;
    else {
      poc = 1;
      kraj = 1000000001;
      while (poc <= kraj) {
        s = (poc + kraj) / 2;
        if (moguce(x, y, p, q, s) == true && moguce(x, y, p, q, s - 1) == false)
          break;
        else {
          if (moguce(x, y, p, q, s) == false)
            poc = s + 1;
          else
            kraj = s - 1;
        }
      }
      a[i] = (long long)q * s - y;
    }
  }
  for (i = 0; i < n; i++) cout << a[i] << endl;
  return 0;
}

```