#include <bits/stdc++.h>
using namespace std;
long long x, y, p, q, a;
int check(int mid) { return ((q * mid - p * mid >= y - x) && (p * mid >= x)); }
long long ter(long long a, long long b) {
  if (a % b == 0)
    return 1LL * a / b;
  else
    return 1LL * a / b + 1;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    if (p == 0) {
      if (x != 0)
        cout << -1 << '\n';
      else
        cout << 0 << '\n';
      continue;
    }
    if (p == q) {
      if (x != y)
        cout << -1 << '\n';
      else
        cout << 0 << '\n';
      continue;
    }
    long long res = 1LL * x * q - 1LL * p * y;
    long long rest = x % p;
    if (rest != 0) rest = p - rest;
    long long res1 = res + 1LL * (q - p) * rest;
    if (res1 < 0) {
      res1 *= -1;
      long long t = rest + ter(res1, (q - p) * p) * p +
                    (-res1 + 1LL * ter(res1, (q - p) * p) * (q - p) * p) / p;
      cout << t << '\n';
    } else
      cout << rest + 1LL * res1 / p << '\n';
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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, a;
int check(int mid) { return ((q * mid - p * mid >= y - x) && (p * mid >= x)); }
long long ter(long long a, long long b) {
  if (a % b == 0)
    return 1LL * a / b;
  else
    return 1LL * a / b + 1;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    if (p == 0) {
      if (x != 0)
        cout << -1 << '\n';
      else
        cout << 0 << '\n';
      continue;
    }
    if (p == q) {
      if (x != y)
        cout << -1 << '\n';
      else
        cout << 0 << '\n';
      continue;
    }
    long long res = 1LL * x * q - 1LL * p * y;
    long long rest = x % p;
    if (rest != 0) rest = p - rest;
    long long res1 = res + 1LL * (q - p) * rest;
    if (res1 < 0) {
      res1 *= -1;
      long long t = rest + ter(res1, (q - p) * p) * p +
                    (-res1 + 1LL * ter(res1, (q - p) * p) * (q - p) * p) / p;
      cout << t << '\n';
    } else
      cout << rest + 1LL * res1 / p << '\n';
  }
  return 0;
}

```