#include <bits/stdc++.h>
using namespace std;
const int INF = int(1e9);
const long long INFll = 1ll * int(1e9) * int(1e9);
const int MOD = 1000000007;
template <typename T, typename Q>
ostream& operator<<(ostream& out, pair<T, Q>& a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename Q>
istream& operator>>(istream& in, pair<T, Q>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (int i = 0; i < a.size(); ++i)
    if (i == a.size() - 1)
      out << a[i];
    else
      out << a[i] << " ";
  return out;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    int x, y, p, q;
    cin >> x >> y >> p >> q;
    long long ans = INFll;
    int l = 1, r = INF;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (q * 1ll * mid < y) {
        l = mid + 1;
        continue;
      }
      long long delta = q * 1ll * mid - y;
      if (p * 1ll * mid - x > delta || p * 1ll * mid - x < 0)
        l = mid + 1;
      else {
        ans = min(ans, delta);
        r = mid - 1;
      }
    }
    if (ans == INFll)
      cout << "-1\n";
    else
      cout << ans << endl;
  }
  if (0)
    cerr << fixed << setprecision(0)
         << "TIME = " << clock() / (long double)CLOCKS_PER_SEC * 1000 << "\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
const int INF = int(1e9);
const long long INFll = 1ll * int(1e9) * int(1e9);
const int MOD = 1000000007;
template <typename T, typename Q>
ostream& operator<<(ostream& out, pair<T, Q>& a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename Q>
istream& operator>>(istream& in, pair<T, Q>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (int i = 0; i < a.size(); ++i) in >> a[i];
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (int i = 0; i < a.size(); ++i)
    if (i == a.size() - 1)
      out << a[i];
    else
      out << a[i] << " ";
  return out;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    int x, y, p, q;
    cin >> x >> y >> p >> q;
    long long ans = INFll;
    int l = 1, r = INF;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (q * 1ll * mid < y) {
        l = mid + 1;
        continue;
      }
      long long delta = q * 1ll * mid - y;
      if (p * 1ll * mid - x > delta || p * 1ll * mid - x < 0)
        l = mid + 1;
      else {
        ans = min(ans, delta);
        r = mid - 1;
      }
    }
    if (ans == INFll)
      cout << "-1\n";
    else
      cout << ans << endl;
  }
  if (0)
    cerr << fixed << setprecision(0)
         << "TIME = " << clock() / (long double)CLOCKS_PER_SEC * 1000 << "\n";
  return 0;
}

```