#include <bits/stdc++.h>
using namespace std;
template <typename T>
void keySort(vector<T> &arr, vector<int> &key) {
  int n = arr.size();
  if (key.size() != n) {
    throw runtime_error("Size of data array not equals to size of key array");
  }
  vector<T> ans(n);
  int ar[n];
  for (int i = 0; i < n; i++) {
    ar[key[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    ans[i] = arr[ar[i]];
  }
  arr = ans;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> p) {
  os << '(' << p.first << ',' << p.second << ')';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> p) {
  os << "{ ";
  for (auto i : p) {
    os << i << ",";
  }
  os << "\b";
  os << "}";
  return os;
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
  in >> p.first >> p.second;
  return in;
}
void solve();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
void solve() {
  long long x, y, a, b;
  cin >> x >> y >> a >> b;
  if (a == 0 && x != 0) {
    std::cout << -1 << endl;
    return;
  }
  if (a == 0 && x == 0) {
    std::cout << 0 << endl;
    return;
  }
  if (a == b && x != y) {
    std::cout << -1 << endl;
    return;
  }
  if (a == b && x == y) {
    std::cout << 0 << endl;
    return;
  }
  long long c = (x + a - 1) / a;
  long long ic = ((y - x) + (b - a) - 1) / (b - a);
  cout << (max(c, ic) * b - y) << endl;
}

### 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>
using namespace std;
template <typename T>
void keySort(vector<T> &arr, vector<int> &key) {
  int n = arr.size();
  if (key.size() != n) {
    throw runtime_error("Size of data array not equals to size of key array");
  }
  vector<T> ans(n);
  int ar[n];
  for (int i = 0; i < n; i++) {
    ar[key[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    ans[i] = arr[ar[i]];
  }
  arr = ans;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> p) {
  os << '(' << p.first << ',' << p.second << ')';
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> p) {
  os << "{ ";
  for (auto i : p) {
    os << i << ",";
  }
  os << "\b";
  os << "}";
  return os;
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
  in >> p.first >> p.second;
  return in;
}
void solve();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
void solve() {
  long long x, y, a, b;
  cin >> x >> y >> a >> b;
  if (a == 0 && x != 0) {
    std::cout << -1 << endl;
    return;
  }
  if (a == 0 && x == 0) {
    std::cout << 0 << endl;
    return;
  }
  if (a == b && x != y) {
    std::cout << -1 << endl;
    return;
  }
  if (a == b && x == y) {
    std::cout << 0 << endl;
    return;
  }
  long long c = (x + a - 1) / a;
  long long ic = ((y - x) + (b - a) - 1) / (b - a);
  cout << (max(c, ic) * b - y) << endl;
}

```