#include <bits/stdc++.h>
using namespace std;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
bool chk(long long p, long long q, long long m, long long a, long long b) {
  long long x = (q * m) - b;
  if (p * m < a || a + x < p * m) return 0;
  return 1;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    int g = gcd(x, y);
    if ((x / g) == p && (y / g) == q) {
      cout << "0\n";
    } else if (p >= q) {
      cout << "-1\n";
    } else {
      long long q1 = 1;
      long long m = 1e18;
      long long q2 = m / q;
      if (y > q) {
        q1 = ceil((y * 1.00) / q);
      }
      while (q1 < q2) {
        long long m = (q1 + q2) / 2;
        if (chk(p, q, m, x, y)) {
          q2 = m;
        } else {
          q1 = m + 1;
        }
      }
      if (chk(p, q, q2, x, y))
        cout << (q * q2 - y) << endl;
      else
        cout << "-1\n";
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
bool chk(long long p, long long q, long long m, long long a, long long b) {
  long long x = (q * m) - b;
  if (p * m < a || a + x < p * m) return 0;
  return 1;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    int g = gcd(x, y);
    if ((x / g) == p && (y / g) == q) {
      cout << "0\n";
    } else if (p >= q) {
      cout << "-1\n";
    } else {
      long long q1 = 1;
      long long m = 1e18;
      long long q2 = m / q;
      if (y > q) {
        q1 = ceil((y * 1.00) / q);
      }
      while (q1 < q2) {
        long long m = (q1 + q2) / 2;
        if (chk(p, q, m, x, y)) {
          q2 = m;
        } else {
          q1 = m + 1;
        }
      }
      if (chk(p, q, q2, x, y))
        cout << (q * q2 - y) << endl;
      else
        cout << "-1\n";
    }
  }
  return 0;
}

```