#include <bits/stdc++.h>
using namespace std;
long long x, y, p, q;
int check(long long mul) {
  long long np = p * mul;
  long long nq = q * mul;
  if (np >= x && nq >= y && (nq - y >= np - x)) return true;
  return false;
}
void solve() {
  cin >> x >> y >> p >> q;
  if (!x && !p) {
    cout << 0 << '\n';
    return;
  }
  if (!!x && !p) {
    cout << -1 << '\n';
    return;
  }
  long long lo = 0, hi = (long long)(1e9) + 5;
  long long res = -1;
  while (lo + 1 < hi) {
    long long mid = lo + (hi - lo) / 2;
    if (check(mid)) {
      res = mid;
      hi = mid;
    } else {
      lo = mid;
    }
  }
  if (res == -1)
    cout << -1 << '\n';
  else {
    long long nq = q * res;
    cout << nq - y << '\n';
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}

### 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 x, y, p, q;
int check(long long mul) {
  long long np = p * mul;
  long long nq = q * mul;
  if (np >= x && nq >= y && (nq - y >= np - x)) return true;
  return false;
}
void solve() {
  cin >> x >> y >> p >> q;
  if (!x && !p) {
    cout << 0 << '\n';
    return;
  }
  if (!!x && !p) {
    cout << -1 << '\n';
    return;
  }
  long long lo = 0, hi = (long long)(1e9) + 5;
  long long res = -1;
  while (lo + 1 < hi) {
    long long mid = lo + (hi - lo) / 2;
    if (check(mid)) {
      res = mid;
      hi = mid;
    } else {
      lo = mid;
    }
  }
  if (res == -1)
    cout << -1 << '\n';
  else {
    long long nq = q * res;
    cout << nq - y << '\n';
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}

```