#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long tc;
  cin >> tc;
  while (tc--) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    if ((p / q) == 1 && x / y != 1) {
      cout << -1 << "\n";
      continue;
    }
    if (p == 0) {
      if (x == 0)
        cout << 0 << "\n";
      else
        cout << -1 << "\n";
      continue;
    }
    long long b = y;
    if (y % q != 0) b = b - y % q + q;
    long long a = (p * b) / q;
    if (a < x) {
      b = (x * q) / p;
      if (b % q != 0) b = b - b % q + q;
    }
    a = p * (b / q);
    long long start = 0, end = 1000000000;
    long long ans = -1;
    while (start <= end) {
      long long mid = (start + end) / 2;
      a = p * ((b + mid * q) / q);
      if ((b + mid * q) - y >= a - x) {
        end = mid - 1;
        ans = mid;
      } else {
        start = mid + 1;
      }
    }
    cout << b + ans * q - y << "\n";
  }
}

### 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;
int32_t main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long tc;
  cin >> tc;
  while (tc--) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    if ((p / q) == 1 && x / y != 1) {
      cout << -1 << "\n";
      continue;
    }
    if (p == 0) {
      if (x == 0)
        cout << 0 << "\n";
      else
        cout << -1 << "\n";
      continue;
    }
    long long b = y;
    if (y % q != 0) b = b - y % q + q;
    long long a = (p * b) / q;
    if (a < x) {
      b = (x * q) / p;
      if (b % q != 0) b = b - b % q + q;
    }
    a = p * (b / q);
    long long start = 0, end = 1000000000;
    long long ans = -1;
    while (start <= end) {
      long long mid = (start + end) / 2;
      a = p * ((b + mid * q) / q);
      if ((b + mid * q) - y >= a - x) {
        end = mid - 1;
        ans = mid;
      } else {
        start = mid + 1;
      }
    }
    cout << b + ans * q - y << "\n";
  }
}

```