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 mod = 1e9 + 7;
const long long inf = 1e18;
int main() {
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    long double x, y, p, q;
    long long a, b, c, d;
    cin >> a >> b >> c >> d;
    x = a, y = b, p = c, q = d;
    if (x == 0 && p == 0) {
      cout << 0 << endl;
      continue;
    }
    if (p == 0 || (p / q == 1 && x / y < 1)) {
      cout << -1 << endl;
      continue;
    }
    if (x / y == p / q) {
      cout << 0 << endl;
      continue;
    }
    if (abs(c * b - a * d) % abs(d - c) == 0 && x / y < p / q) {
      cout << abs(c * b - a * d) / abs(d - c) << endl;
      continue;
    }
    long long cur = max((a + c - 1) / c, (b + d - 1) / d);
    cur = max(cur, (b - a + d - c) / (d - c));
    cout << cur * d - b << endl;
  }
}
