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>
using namespace std;
const int size = 1e5 + 7;
const int maxm = 2e9 + 7;
long long p, q, x, y;
bool check(int mid) {
  return (p * mid - x >= 0 && q * mid - y >= 0 && (q - p) * mid - (y - x) >= 0);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  long long l, r, mid;
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    l = 0, r = 2e9;
    while (l < r) {
      mid = l + r >> 1;
      if (check(mid))
        r = mid;
      else
        l = mid + 1;
    }
    if (r == 2e9)
      cout << -1 << endl;
    else
      cout << q * r - y << endl;
  }
  return 0;
}
