#include <bits/stdc++.h>
using namespace std;
inline long long ll_ceil_of_div(long long a, long long b) {
  if (a % b == 0LL)
    return a / b;
  else
    return 1LL + (a / b);
}
inline long long ll_ceil_without_div(long long a, long long b) {
  long long rem = a % b;
  if (rem == 0LL)
    return a;
  else
    return a + b - rem;
}
int main(int argc, char* argv[]) {
  cout.sync_with_stdio(false);
  int TEST_NUM;
  scanf("%d", &TEST_NUM);
  for (int the_test = 0; the_test < TEST_NUM; the_test++) {
    int x_int, y_int, p_int, q_int;
    scanf("%d %d %d %d", &x_int, &y_int, &p_int, &q_int);
    long long x = x_int;
    long long y = y_int;
    long long p = p_int;
    long long q = q_int;
    if (p == 0) {
      if (x == 0)
        cout << "0\n";
      else
        cout << "-1\n";
    } else if (p == q) {
      if (x == y)
        cout << "0\n";
      else
        cout << "-1\n";
    } else if (x * q == y * p) {
      cout << "0\n";
    } else {
      long long res_1, res_2;
      if (x * q < y * p) {
        res_1 = ll_ceil_of_div(p * y - x * q, q - p);
      } else {
        res_1 = ll_ceil_of_div(x * q - p * y, p);
      }
      res_2 = ll_ceil_without_div(y + res_1, q) - y;
      cout << res_2 << "\n";
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
inline long long ll_ceil_of_div(long long a, long long b) {
  if (a % b == 0LL)
    return a / b;
  else
    return 1LL + (a / b);
}
inline long long ll_ceil_without_div(long long a, long long b) {
  long long rem = a % b;
  if (rem == 0LL)
    return a;
  else
    return a + b - rem;
}
int main(int argc, char* argv[]) {
  cout.sync_with_stdio(false);
  int TEST_NUM;
  scanf("%d", &TEST_NUM);
  for (int the_test = 0; the_test < TEST_NUM; the_test++) {
    int x_int, y_int, p_int, q_int;
    scanf("%d %d %d %d", &x_int, &y_int, &p_int, &q_int);
    long long x = x_int;
    long long y = y_int;
    long long p = p_int;
    long long q = q_int;
    if (p == 0) {
      if (x == 0)
        cout << "0\n";
      else
        cout << "-1\n";
    } else if (p == q) {
      if (x == y)
        cout << "0\n";
      else
        cout << "-1\n";
    } else if (x * q == y * p) {
      cout << "0\n";
    } else {
      long long res_1, res_2;
      if (x * q < y * p) {
        res_1 = ll_ceil_of_div(p * y - x * q, q - p);
      } else {
        res_1 = ll_ceil_of_div(x * q - p * y, p);
      }
      res_2 = ll_ceil_without_div(y + res_1, q) - y;
      cout << res_2 << "\n";
    }
  }
  return 0;
}

```