#include <bits/stdc++.h>
using namespace std;
const long double inf = 1e18 + 7;
const long double eps = 1e-18;
long long gcd(long long a, long long b) {
  while (a || b) a %= b, swap(a, b);
  return a + b;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long t;
  cin >> t;
  for (long long tt = 0; tt < t; ++tt) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    long long ok = y - x;
    long long wa = y - ok;
    long long nok = q - p;
    long long nwa = q - nok;
    if (nok == nwa) {
      cout << abs(ok - wa) << endl;
      continue;
    }
    if (nok == 0) {
      if (ok != 0) {
        cout << -1 << endl;
        continue;
      } else {
        cout << 0 << endl;
        continue;
      }
    }
    if (nwa == 0) {
      if (wa != 0) {
        cout << -1 << endl;
        continue;
      } else {
        cout << 0 << endl;
        continue;
      }
    }
    if (nok < nwa) {
      long long xx = ok + nok - 1;
      xx /= nok;
      xx *= nok;
      long long l = -1;
      long long r = 1e10;
      while (l != r - 1) {
        long long m = (l + r) / 2ll;
        long long lx = xx + nok * m;
        long long rx = lx / nok * nwa;
        if (rx >= wa)
          r = m;
        else
          l = m;
      }
      long long lx = xx + nok * r;
      long long rx = lx / nok * nwa;
      long long ans = lx - ok;
      ans += rx - wa;
      cout << ans << endl;
    } else {
      long long xx = wa + nwa - 1;
      xx /= nwa;
      xx *= nwa;
      long long l = -1;
      long long r = 1e10;
      while (l != r - 1) {
        long long m = (l + r) / 2;
        long long lx = xx + m * nwa;
        long long rx = lx / nwa * nok;
        if (rx >= ok)
          r = m;
        else
          l = m;
      }
      long long lx = xx + r * nwa;
      long long rx = lx / nwa * nok;
      long long ans = (lx - wa) + (rx - ok);
      cout << ans << endl;
    }
  }
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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;
const long double inf = 1e18 + 7;
const long double eps = 1e-18;
long long gcd(long long a, long long b) {
  while (a || b) a %= b, swap(a, b);
  return a + b;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long t;
  cin >> t;
  for (long long tt = 0; tt < t; ++tt) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    long long ok = y - x;
    long long wa = y - ok;
    long long nok = q - p;
    long long nwa = q - nok;
    if (nok == nwa) {
      cout << abs(ok - wa) << endl;
      continue;
    }
    if (nok == 0) {
      if (ok != 0) {
        cout << -1 << endl;
        continue;
      } else {
        cout << 0 << endl;
        continue;
      }
    }
    if (nwa == 0) {
      if (wa != 0) {
        cout << -1 << endl;
        continue;
      } else {
        cout << 0 << endl;
        continue;
      }
    }
    if (nok < nwa) {
      long long xx = ok + nok - 1;
      xx /= nok;
      xx *= nok;
      long long l = -1;
      long long r = 1e10;
      while (l != r - 1) {
        long long m = (l + r) / 2ll;
        long long lx = xx + nok * m;
        long long rx = lx / nok * nwa;
        if (rx >= wa)
          r = m;
        else
          l = m;
      }
      long long lx = xx + nok * r;
      long long rx = lx / nok * nwa;
      long long ans = lx - ok;
      ans += rx - wa;
      cout << ans << endl;
    } else {
      long long xx = wa + nwa - 1;
      xx /= nwa;
      xx *= nwa;
      long long l = -1;
      long long r = 1e10;
      while (l != r - 1) {
        long long m = (l + r) / 2;
        long long lx = xx + m * nwa;
        long long rx = lx / nwa * nok;
        if (rx >= ok)
          r = m;
        else
          l = m;
      }
      long long lx = xx + r * nwa;
      long long rx = lx / nwa * nok;
      long long ans = (lx - wa) + (rx - ok);
      cout << ans << endl;
    }
  }
  return 0;
}

```