#include <bits/stdc++.h>
using namespace std;
double pi = 3.141592653589793238462;
string numkey = "0123456789";
string uni = "abcdefghijklmnopqrstuvwxyz";
long long int dx[4] = {0, 0, 1, -1};
long long int dy[4] = {1, -1, 0, 0};
long long int power(long long int x, long long int y) {
  if (y == 0) return 1;
  long long int res = power(x, y / 2);
  if (y % 2 == 0) return res * res;
  return res * res * x;
}
bool check(long long int mid, long long int x, long long int y, long long int p,
           long long int q) {
  if ((p * mid >= x) && (q - p) * mid >= (y - x)) return true;
  return false;
}
signed main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int x, y, p, q;
    cin >> x >> y >> p >> q;
    long long int l = -1, r = 1000000000, ans = -1;
    if (check(r, x, y, p, q) == false) {
      cout << "-1\n";
      continue;
    }
    while (r - l > 1) {
      long long int mid = (l + r) / 2;
      if (check(mid, x, y, p, q)) {
        ans = mid;
        r = mid;
      } else
        l = mid;
    }
    cout << (r * q - y) << "\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;
double pi = 3.141592653589793238462;
string numkey = "0123456789";
string uni = "abcdefghijklmnopqrstuvwxyz";
long long int dx[4] = {0, 0, 1, -1};
long long int dy[4] = {1, -1, 0, 0};
long long int power(long long int x, long long int y) {
  if (y == 0) return 1;
  long long int res = power(x, y / 2);
  if (y % 2 == 0) return res * res;
  return res * res * x;
}
bool check(long long int mid, long long int x, long long int y, long long int p,
           long long int q) {
  if ((p * mid >= x) && (q - p) * mid >= (y - x)) return true;
  return false;
}
signed main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int x, y, p, q;
    cin >> x >> y >> p >> q;
    long long int l = -1, r = 1000000000, ans = -1;
    if (check(r, x, y, p, q) == false) {
      cout << "-1\n";
      continue;
    }
    while (r - l > 1) {
      long long int mid = (l + r) / 2;
      if (check(mid, x, y, p, q)) {
        ans = mid;
        r = mid;
      } else
        l = mid;
    }
    cout << (r * q - y) << "\n";
  }
  return 0;
}

```