#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int rez, t, p, q, a, b, x, y, k;
  cin >> t;
  while (t--) {
    k = 0;
    cin >> x >> y >> p >> q;
    if (q == 1 or p == 0) {
      if (p == 1 and q == 1) {
        if (x < y) cout << -1 << '\n';
        if (x == y) cout << 0 << '\n';
      }
      if (p == 0) {
        if (x > 0) cout << -1 << '\n';
        if (x == 0) cout << 0 << '\n';
      }
    } else {
      if (x % p == 0 and k < x / p) k = x / p;
      if (x % p != 0 and k <= x / p) k = x / p + 1;
      if (y % q == 0 and k < y / q) k = y / q;
      if (y % q != 0 and k <= y / q) k = y / q + 1;
      a = y - x;
      b = q - p;
      if (a % b == 0 and k < a / b) k = a / b;
      if (a % b != 0 and k <= a / b) k = a / b + 1;
      rez = q * k - y;
      cout << rez << '\n';
    }
  }
  return 0;
}

### Prompt

Generate 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 main() {
  long long int rez, t, p, q, a, b, x, y, k;
  cin >> t;
  while (t--) {
    k = 0;
    cin >> x >> y >> p >> q;
    if (q == 1 or p == 0) {
      if (p == 1 and q == 1) {
        if (x < y) cout << -1 << '\n';
        if (x == y) cout << 0 << '\n';
      }
      if (p == 0) {
        if (x > 0) cout << -1 << '\n';
        if (x == 0) cout << 0 << '\n';
      }
    } else {
      if (x % p == 0 and k < x / p) k = x / p;
      if (x % p != 0 and k <= x / p) k = x / p + 1;
      if (y % q == 0 and k < y / q) k = y / q;
      if (y % q != 0 and k <= y / q) k = y / q + 1;
      a = y - x;
      b = q - p;
      if (a % b == 0 and k < a / b) k = a / b;
      if (a % b != 0 and k <= a / b) k = a / b + 1;
      rez = q * k - y;
      cout << rez << '\n';
    }
  }
  return 0;
}

```