#include <bits/stdc++.h>
using namespace std;
long long n, a, b, p, q;
long long gcd(long long a, long long b) {
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
bool check(long long x) { return x * p - a >= 0 && x * q - b >= x * p - a; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a >> b >> p >> q;
    long long first = 0, last = 2e9, mid;
    if (a / gcd(a, b) == p && b / gcd(a, b) == q) {
      cout << "0" << endl;
      continue;
    }
    if ((p == 1 && q == 1) || p == 0) {
      cout << "-1" << endl;
      continue;
    }
    while (first < last) {
      mid = first + (last - first) / 2;
      if (check(mid) == true) {
        last = mid;
      } else
        first = mid + 1;
    }
    if (first != 0 && last != 2e9)
      cout << first * q - b << endl;
    else
      cout << "-1" << endl;
  }
  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;
long long n, a, b, p, q;
long long gcd(long long a, long long b) {
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
bool check(long long x) { return x * p - a >= 0 && x * q - b >= x * p - a; }
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a >> b >> p >> q;
    long long first = 0, last = 2e9, mid;
    if (a / gcd(a, b) == p && b / gcd(a, b) == q) {
      cout << "0" << endl;
      continue;
    }
    if ((p == 1 && q == 1) || p == 0) {
      cout << "-1" << endl;
      continue;
    }
    while (first < last) {
      mid = first + (last - first) / 2;
      if (check(mid) == true) {
        last = mid;
      } else
        first = mid + 1;
    }
    if (first != 0 && last != 2e9)
      cout << first * q - b << endl;
    else
      cout << "-1" << endl;
  }
  return 0;
}

```