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 pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long t, x, y, p, q;
long long binary(long long min, long long max) {
  if (min >= max) return max;
  long long mid = (min + max) / 2;
  if (p * mid - x <= q * mid - y && p * mid - x >= 0)
    return binary(min, mid);
  else
    return binary(mid + 1, max);
}
int main() {
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    if (p == 0 && x == 0) {
      cout << "0\n";
      continue;
    }
    if (p == 0 && x != 0) {
      cout << "-1\n";
      continue;
    }
    if (y == 0) {
      cout << q << "\n";
      continue;
    }
    if (p == q && x == y) {
      cout << "0\n";
      continue;
    }
    if (p == q) {
      cout << "-1\n";
      continue;
    }
    long long ans = binary(1, 1e9);
    cout << q * ans - y << "\n";
  }
  return 0;
}
