#include <bits/stdc++.h>
using namespace std;
long long t, x, y, p, q, ans;
long long bp(int l, int r) {
  int m;
  while (r - l > 1) {
    m = (r + l) / 2;
    if (((x + ans + m * p) / p * q - y) >= ans + m * p)
      r = m;
    else
      l = m;
  }
  return l;
}
long long nod(long long a, long long b) {
  if (b > a) swap(a, b);
  if (b != 0 && a != b)
    return nod(a % b, b);
  else
    return a;
}
int main() {
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> x >> y >> p >> q;
    if (x != y && p == q) {
      cout << -1 << endl;
      continue;
    }
    if (x != 0 && p == 0) {
      cout << -1 << endl;
      continue;
    }
    if (x == 0 && p == 0) {
      cout << 0 << endl;
      continue;
    }
    long long m = nod(p, q);
    q = q / m;
    p = p / m;
    ans = max((long long)0, (y * p + q - 1) / q - x);
    long long a = (p - x % p) % p;
    ans = ans / p * p + a;
    m = bp(0, 1e9);
    if ((x + ans + m * p) / p * q - y < ans + m * p) m++;
    cout << (x + ans + m * p) / p * q - y << endl;
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to 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 t, x, y, p, q, ans;
long long bp(int l, int r) {
  int m;
  while (r - l > 1) {
    m = (r + l) / 2;
    if (((x + ans + m * p) / p * q - y) >= ans + m * p)
      r = m;
    else
      l = m;
  }
  return l;
}
long long nod(long long a, long long b) {
  if (b > a) swap(a, b);
  if (b != 0 && a != b)
    return nod(a % b, b);
  else
    return a;
}
int main() {
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> x >> y >> p >> q;
    if (x != y && p == q) {
      cout << -1 << endl;
      continue;
    }
    if (x != 0 && p == 0) {
      cout << -1 << endl;
      continue;
    }
    if (x == 0 && p == 0) {
      cout << 0 << endl;
      continue;
    }
    long long m = nod(p, q);
    q = q / m;
    p = p / m;
    ans = max((long long)0, (y * p + q - 1) / q - x);
    long long a = (p - x % p) % p;
    ans = ans / p * p + a;
    m = bp(0, 1e9);
    if ((x + ans + m * p) / p * q - y < ans + m * p) m++;
    cout << (x + ans + m * p) / p * q - y << endl;
  }
  return 0;
}

```