#include <bits/stdc++.h>
using namespace std;
long long func(long long x, long long y, long long p, long long q) {
  long long temp, diff, base = 1, a, b, ans;
  if (y > q) {
    base = y / q;
    if (y % q) base++;
  }
  a = p * base;
  diff = q * base - y;
  if (x > a) {
    x = x - a;
    temp = x / p;
    if (x % p) temp++;
    base += temp;
    ans = q * base - y;
    return ans;
  } else if (x + diff < a) {
    x += diff;
    diff = a - x;
    b = q - p;
    temp = diff / b;
    if (diff % b) temp++;
    base += temp;
    ans = q * base - y;
    return ans;
  }
  return diff;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, x, y, p, q;
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    if (p == 0) {
      if (x == 0)
        cout << "0";
      else
        cout << "-1";
    } else if (p == q) {
      if (x == y)
        cout << "0";
      else
        cout << "-1";
    } else
      cout << func(x, y, p, q);
    cout << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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 func(long long x, long long y, long long p, long long q) {
  long long temp, diff, base = 1, a, b, ans;
  if (y > q) {
    base = y / q;
    if (y % q) base++;
  }
  a = p * base;
  diff = q * base - y;
  if (x > a) {
    x = x - a;
    temp = x / p;
    if (x % p) temp++;
    base += temp;
    ans = q * base - y;
    return ans;
  } else if (x + diff < a) {
    x += diff;
    diff = a - x;
    b = q - p;
    temp = diff / b;
    if (diff % b) temp++;
    base += temp;
    ans = q * base - y;
    return ans;
  }
  return diff;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, x, y, p, q;
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    if (p == 0) {
      if (x == 0)
        cout << "0";
      else
        cout << "-1";
    } else if (p == q) {
      if (x == y)
        cout << "0";
      else
        cout << "-1";
    } else
      cout << func(x, y, p, q);
    cout << "\n";
  }
  return 0;
}

```