#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T, class S>
ostream &operator<<(ostream &o, const pair<T, S> &p) {
  return o << '(' << p.first << ", " << p.second << ')';
}
template <template <class, class...> class T, class... A>
typename enable_if<!is_same<T<A...>, string>(), ostream &>::type operator<<(
    ostream &o, T<A...> V) {
  o << '[';
  for (auto a : V) o << a << ", ";
  return o << ']';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int tc;
  cin >> tc;
  while (tc--) {
    long long int x;
    cin >> x;
    long long int y;
    cin >> y;
    long long int p;
    cin >> p;
    long long int q;
    cin >> q;
    if (!p) {
      cout << (x ? "-1\n" : "0\n");
    } else if (p == 1 && q == 1) {
      cout << (x == y ? "0\n" : "-1\n");
    } else {
      long long int mindenom = max(y, (x * q + p - 1) / p);
      mindenom = max(mindenom, y + (p * y - x * q + q - p - 1) / (q - p));
      if (mindenom % q) mindenom += q - mindenom % q;
      cout << mindenom - y << '\n';
    }
  }
}

### Prompt

Create a solution in Cpp for 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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T, class S>
ostream &operator<<(ostream &o, const pair<T, S> &p) {
  return o << '(' << p.first << ", " << p.second << ')';
}
template <template <class, class...> class T, class... A>
typename enable_if<!is_same<T<A...>, string>(), ostream &>::type operator<<(
    ostream &o, T<A...> V) {
  o << '[';
  for (auto a : V) o << a << ", ";
  return o << ']';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int tc;
  cin >> tc;
  while (tc--) {
    long long int x;
    cin >> x;
    long long int y;
    cin >> y;
    long long int p;
    cin >> p;
    long long int q;
    cin >> q;
    if (!p) {
      cout << (x ? "-1\n" : "0\n");
    } else if (p == 1 && q == 1) {
      cout << (x == y ? "0\n" : "-1\n");
    } else {
      long long int mindenom = max(y, (x * q + p - 1) / p);
      mindenom = max(mindenom, y + (p * y - x * q + q - p - 1) / (q - p));
      if (mindenom % q) mindenom += q - mindenom % q;
      cout << mindenom - y << '\n';
    }
  }
}

```