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>
#pragma GCC optimize("Ofast")
using namespace std;
int32_t main() {
  int t;
  cin >> t;
  for (int z = 0; z < t; z += 1) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    if (p == 0) {
      if (x == 0)
        cout << 0 << '\n';
      else
        cout << -1 << '\n';
      continue;
    }
    long long left = 0;
    long long right = 1e9;
    long long ans = -1;
    long long cur_mod = abs(x - y);
    while (left <= right) {
      long long mid = (left + right) / 2;
      long long zc = (mid * q);
      if (zc < y) {
        left = mid + 1;
        continue;
      }
      long long wow = p * (mid);
      if (wow < x) {
        left = mid + 1;
        continue;
      }
      if (abs(wow - zc) < cur_mod) {
        left = mid + 1;
        continue;
      } else {
        ans = mid;
        right = mid - 1;
      }
    }
    if (ans < 0)
      cout << ans << '\n';
    else
      cout << (q * ans) - y << '\n';
  }
}
