#include <bits/stdc++.h>
using namespace std;
bool function(int x, int y) { return (x > y); }
int main() {
  long long int t, x, y, p, q, c1, c2, mult, flag, temp;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> x >> y >> p >> q;
    if (p == 0) {
      if (x == 0) {
        cout << "0" << endl;
      } else
        cout << "-1" << endl;
      continue;
    }
    if (q >= y) {
      mult = 1;
      flag = x / p;
      if (flag * p < x) flag++;
      if (flag > mult) mult = flag;
      long long int val2 = mult * q - y;
      if ((mult * p - x) > (mult * q - y)) {
        c2 = mult * p - x;
        if (p == q)
          cout << "-1" << endl;
        else {
          temp = c2 - val2;
          c2 = (c2 - val2) / (q - p);
          if (c2 * (q - p) < temp) c2++;
          mult = c2 + mult;
          c1 = (mult * q - y);
          cout << c1 << endl;
        }
      } else
        cout << val2 << endl;
    } else {
      mult = y / q;
      if (mult * q < y) mult++;
      flag = x / p;
      if (flag * p < x) flag++;
      if (flag > mult) mult = flag;
      flag = 0;
      long long int val2 = mult * q - y;
      if ((mult * p - x) > (mult * q - y)) {
        c2 = mult * p - x;
        if (p == q)
          cout << "-1" << endl;
        else {
          temp = c2 - val2;
          c2 = (c2 - val2) / (q - p);
          if (c2 * (q - p) < temp) c2++;
          mult = c2 + mult;
          c1 = (mult * q - y);
          cout << c1 << endl;
        }
      } else
        cout << val2 << 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;
bool function(int x, int y) { return (x > y); }
int main() {
  long long int t, x, y, p, q, c1, c2, mult, flag, temp;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> x >> y >> p >> q;
    if (p == 0) {
      if (x == 0) {
        cout << "0" << endl;
      } else
        cout << "-1" << endl;
      continue;
    }
    if (q >= y) {
      mult = 1;
      flag = x / p;
      if (flag * p < x) flag++;
      if (flag > mult) mult = flag;
      long long int val2 = mult * q - y;
      if ((mult * p - x) > (mult * q - y)) {
        c2 = mult * p - x;
        if (p == q)
          cout << "-1" << endl;
        else {
          temp = c2 - val2;
          c2 = (c2 - val2) / (q - p);
          if (c2 * (q - p) < temp) c2++;
          mult = c2 + mult;
          c1 = (mult * q - y);
          cout << c1 << endl;
        }
      } else
        cout << val2 << endl;
    } else {
      mult = y / q;
      if (mult * q < y) mult++;
      flag = x / p;
      if (flag * p < x) flag++;
      if (flag > mult) mult = flag;
      flag = 0;
      long long int val2 = mult * q - y;
      if ((mult * p - x) > (mult * q - y)) {
        c2 = mult * p - x;
        if (p == q)
          cout << "-1" << endl;
        else {
          temp = c2 - val2;
          c2 = (c2 - val2) / (q - p);
          if (c2 * (q - p) < temp) c2++;
          mult = c2 + mult;
          c1 = (mult * q - y);
          cout << c1 << endl;
        }
      } else
        cout << val2 << endl;
    }
  }
  return 0;
}

```