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.
#include <bits/stdc++.h>
const double PI =
    3.141592653589793238462643383279502884197169399375105820974944;
using namespace std;
long long x, y, p, q;
bool func(long long mid) {
  long long temp = p * mid;
  long long temp1 = q * mid;
  return temp >= x && temp1 >= y && (abs(temp - x) <= abs(temp1 - y));
}
void solve() {
  int t;
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    long long low = 1, high = 1e9;
    while (low < high) {
      long long mid = (low + high) / 2;
      if (func(mid)) {
        high = mid;
      } else {
        low = mid + 1;
      }
    }
    if (func(low))
      cout << q * low - y << "\n";
    else
      cout << "-1"
           << "\n";
  }
}
int main() {
  bool testing = false;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (testing) {
    freopen("test.txt", "rt", stdin);
    int start = clock();
    solve();
    int end = clock();
    cout << "time: " << (end - start) / (double)(CLOCKS_PER_SEC)*1000
         << " milliseconds\n";
  } else {
    solve();
  }
}
