#include <bits/stdc++.h>
using namespace std;
int t;
double x, y, p, q;
long long ans = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    if (p == 0) {
      if (x == 0)
        ans = 0;
      else
        ans = -1;
    } else if (p == q) {
      if (x != y) {
        ans = -1;
      } else
        ans = 0;
    } else {
      long long st = 0, en = 2e18;
      long long mid = (st + en) / 2;
      while (st != en) {
        if ((x + (double)mid) * (q - p) >= (y - x) * p)
          en = mid;
        else
          st = mid + 1;
        mid = (st + en) / 2;
      }
      ans = mid;
      if ((x + (double)mid) * (q - p) > (y - x) * p) {
        long long px = x, py = y, pq = q, pp = p;
        px += mid;
        py += mid;
        if (px % pp != 0) {
          int pl = px % pp;
          px += pp - pl;
          py += pp - pl;
          ans += pp - pl;
        }
        ans += ((px / pp) * pq) - py;
      }
    }
    cout << ans << endl;
  }
}

### Prompt

Your task is to create 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;
int t;
double x, y, p, q;
long long ans = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    if (p == 0) {
      if (x == 0)
        ans = 0;
      else
        ans = -1;
    } else if (p == q) {
      if (x != y) {
        ans = -1;
      } else
        ans = 0;
    } else {
      long long st = 0, en = 2e18;
      long long mid = (st + en) / 2;
      while (st != en) {
        if ((x + (double)mid) * (q - p) >= (y - x) * p)
          en = mid;
        else
          st = mid + 1;
        mid = (st + en) / 2;
      }
      ans = mid;
      if ((x + (double)mid) * (q - p) > (y - x) * p) {
        long long px = x, py = y, pq = q, pp = p;
        px += mid;
        py += mid;
        if (px % pp != 0) {
          int pl = px % pp;
          px += pp - pl;
          py += pp - pl;
          ans += pp - pl;
        }
        ans += ((px / pp) * pq) - py;
      }
    }
    cout << ans << endl;
  }
}

```