#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000000 + 7;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return gcd(b, a % b);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    long long div = gcd(x, y);
    long long x1 = x / div;
    long long y1 = y / div;
    if (x1 == p && y1 == q) {
      cout << 0 << endl;
      continue;
    }
    if (p == 1 && q == 1) {
      cout << -1 << endl;
      continue;
    }
    long long st = 1, ed = 10000000000, mid, prevmid = -1;
    while (st <= ed) {
      mid = (st + ed) / 2;
      if (mid == prevmid) break;
      long long y1 = q * mid, x1 = p * mid;
      if ((x1 < x) || (y1 < y) || ((p * mid - x) > (q * mid - y)))
        st = mid + 1;
      else
        ed = mid;
      prevmid = mid;
    }
    if (st > ed) {
      cout << -1 << endl;
      continue;
    } else {
      long long ans = q * mid - y;
      cout << ans << endl;
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000000 + 7;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return gcd(b, a % b);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    long long div = gcd(x, y);
    long long x1 = x / div;
    long long y1 = y / div;
    if (x1 == p && y1 == q) {
      cout << 0 << endl;
      continue;
    }
    if (p == 1 && q == 1) {
      cout << -1 << endl;
      continue;
    }
    long long st = 1, ed = 10000000000, mid, prevmid = -1;
    while (st <= ed) {
      mid = (st + ed) / 2;
      if (mid == prevmid) break;
      long long y1 = q * mid, x1 = p * mid;
      if ((x1 < x) || (y1 < y) || ((p * mid - x) > (q * mid - y)))
        st = mid + 1;
      else
        ed = mid;
      prevmid = mid;
    }
    if (st > ed) {
      cout << -1 << endl;
      continue;
    } else {
      long long ans = q * mid - y;
      cout << ans << endl;
    }
  }
  return 0;
}

```