#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF = ~(1 << 31);
bool isOk(ll x, ll y, ll p, ll q, ll t) {
  return (p * t >= x && (q - p) * t >= y - x);
}
ll bs(ll x, ll y, ll p, ll q) {
  ll st = 1, dr = INF, med, last;
  if (!isOk(x, y, p, q, INF)) return -1;
  while (st <= dr) {
    med = (st + dr) / 2;
    if (isOk(x, y, p, q, med)) {
      dr = med - 1;
      last = med;
    } else
      st = med + 1;
  }
  return last * q - y;
}
int main() {
  {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
  };
  int t;
  cin >> t;
  ll x, y, p, q;
  for (int i = 1; i <= t; i++) {
    cin >> x >> y >> p >> q;
    cout << bs(x, y, p, q) << '\n';
  }
  { return 0; };
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
using ll = long long;
const int INF = ~(1 << 31);
bool isOk(ll x, ll y, ll p, ll q, ll t) {
  return (p * t >= x && (q - p) * t >= y - x);
}
ll bs(ll x, ll y, ll p, ll q) {
  ll st = 1, dr = INF, med, last;
  if (!isOk(x, y, p, q, INF)) return -1;
  while (st <= dr) {
    med = (st + dr) / 2;
    if (isOk(x, y, p, q, med)) {
      dr = med - 1;
      last = med;
    } else
      st = med + 1;
  }
  return last * q - y;
}
int main() {
  {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
  };
  int t;
  cin >> t;
  ll x, y, p, q;
  for (int i = 1; i <= t; i++) {
    cin >> x >> y >> p >> q;
    cout << bs(x, y, p, q) << '\n';
  }
  { return 0; };
}

```