#include <bits/stdc++.h>
using namespace std;
long long p, q;
long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b != 0) {
    if (b % 2 == 1) ans = (ans * a) % q;
    a = a * a % q;
    b = b / 2;
  }
  return ans;
}
long long eula(long long x) {
  long long ans = x;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      ans = ans - ans / i;
      while (x % i == 0) x /= i;
    }
  }
  if (x != 1) ans = ans - ans / x;
  return ans;
}
void linear_equation(long long a, long long b, long long c, long long &x,
                     long long &y) {
  x = (ksm(a, eula(b) - 1)) * (c % q) % q;
  y = (c - a * x) / b;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long x, y;
    cin >> x >> y >> p >> q;
    if (p == 0 && x != 0) {
      puts("-1");
      continue;
    }
    if (p == 0 && x == 0) {
      puts("0");
      continue;
    }
    if (p == q && x < y) {
      puts("-1");
      continue;
    }
    if (p == q && x == y) {
      puts("0");
      continue;
    }
    long long l1, l2;
    linear_equation(-p, q, -q * x + p * y, l1, l2);
    long long ans1 = (l1 % q + q) % q;
    long long ans2 = (l2 % p + p) % p;
    long long tt = max(((ans1 - l1) / q), ((ans2 - l2) / p));
    ans1 = l1 + q * tt;
    ans2 = l2 + p * tt;
    long long d = ans2 - ans1;
    if (d <= 0)
      cout << ans1 << endl;
    else
      cout << ans1 + ((d + ((q - p - 1))) / (q - p)) * q << 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;
long long p, q;
long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b != 0) {
    if (b % 2 == 1) ans = (ans * a) % q;
    a = a * a % q;
    b = b / 2;
  }
  return ans;
}
long long eula(long long x) {
  long long ans = x;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      ans = ans - ans / i;
      while (x % i == 0) x /= i;
    }
  }
  if (x != 1) ans = ans - ans / x;
  return ans;
}
void linear_equation(long long a, long long b, long long c, long long &x,
                     long long &y) {
  x = (ksm(a, eula(b) - 1)) * (c % q) % q;
  y = (c - a * x) / b;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long x, y;
    cin >> x >> y >> p >> q;
    if (p == 0 && x != 0) {
      puts("-1");
      continue;
    }
    if (p == 0 && x == 0) {
      puts("0");
      continue;
    }
    if (p == q && x < y) {
      puts("-1");
      continue;
    }
    if (p == q && x == y) {
      puts("0");
      continue;
    }
    long long l1, l2;
    linear_equation(-p, q, -q * x + p * y, l1, l2);
    long long ans1 = (l1 % q + q) % q;
    long long ans2 = (l2 % p + p) % p;
    long long tt = max(((ans1 - l1) / q), ((ans2 - l2) / p));
    ans1 = l1 + q * tt;
    ans2 = l2 + p * tt;
    long long d = ans2 - ans1;
    if (d <= 0)
      cout << ans1 << endl;
    else
      cout << ans1 + ((d + ((q - p - 1))) / (q - p)) * q << endl;
  }
  return 0;
}

```