#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int powm(long long int x, long long int y,
                   long long int m = 1000000007) {
  x = x % m;
  long long int res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long int modInverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int T;
  cin >> T;
  while (T--) {
    long long int x, y, p, q;
    cin >> x >> y >> p >> q;
    long long int a;
    if (p == 0)
      a = 1e18;
    else
      a = (x + p - 1) / p;
    long long int b;
    if (q == 0)
      b = 1e18;
    else
      b = (y + q - 1) / q;
    if (p == 0) {
      if (x == 0)
        cout << 0 << "\n";
      else
        cout << -1 << "\n";
      continue;
    }
    if ((q == p)) {
      if (x != y)
        cout << -1 << "\n";
      else
        cout << 0 << "\n";
      continue;
    }
    long long int j = (x + p - 1) / p;
    long long int k = (y + q - 1) / q;
    long long int i = (y - x + q - p - 1) / (q - p);
    i = max(i, j);
    i = max(i, k);
    p *= i;
    q *= i;
    cout << q - y << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int powm(long long int x, long long int y,
                   long long int m = 1000000007) {
  x = x % m;
  long long int res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long int modInverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int T;
  cin >> T;
  while (T--) {
    long long int x, y, p, q;
    cin >> x >> y >> p >> q;
    long long int a;
    if (p == 0)
      a = 1e18;
    else
      a = (x + p - 1) / p;
    long long int b;
    if (q == 0)
      b = 1e18;
    else
      b = (y + q - 1) / q;
    if (p == 0) {
      if (x == 0)
        cout << 0 << "\n";
      else
        cout << -1 << "\n";
      continue;
    }
    if ((q == p)) {
      if (x != y)
        cout << -1 << "\n";
      else
        cout << 0 << "\n";
      continue;
    }
    long long int j = (x + p - 1) / p;
    long long int k = (y + q - 1) / q;
    long long int i = (y - x + q - p - 1) / (q - p);
    i = max(i, j);
    i = max(i, k);
    p *= i;
    q *= i;
    cout << q - y << "\n";
  }
  return 0;
}

```