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;
long long flr(long long x, long long p) {
  if (x % p == 0) return x;
  return x + p - x % p;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long tests, a, b, c, d, x, y, t, p, q;
  cin >> tests;
  while (tests--) {
    cin >> a >> b >> c >> d;
    if (c == 0 && d == 1) {
      if (a != 0) {
        cout << -1 << endl;
        continue;
      }
      cout << 0 << endl;
      continue;
    }
    if (c == d) {
      if (a < b)
        cout << -1 << endl;
      else
        cout << 0 << endl;
    } else {
      y = d - c;
      x = b - a;
      t = flr(x, y) / y;
      p = t * c;
      q = t * d;
      if (p < a) {
        p = flr(a, c);
        q = p / c * d;
      }
      if (q < b) {
        q = flr(b, d);
        p = q / d * c;
      }
      cout << q - b << endl;
    }
  }
}
