#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
const int M = 1000000007;
int n;
long long x, y, p, q;
bool ok(long long k) {
  long long pp = p * k, qq = q * k;
  return pp >= x && qq >= y && qq - y >= pp - x;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      cin >> x >> y >> p >> q;
      if (p == q && x != y) {
        puts("-1");
        continue;
      }
      if (x && p == 0) {
        puts("-1");
        continue;
      }
      if (x * q == y * p) {
        puts("0");
        continue;
      }
      long long l = 1, r = 2e9;
      while (l <= r) {
        long long mid = (l + r) >> 1;
        if (ok(mid)) {
          r = mid - 1;
        } else {
          l = mid + 1;
        }
      }
      cout << q * l - y << '\n';
    }
  }
}

### 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 int N = 100005;
const int M = 1000000007;
int n;
long long x, y, p, q;
bool ok(long long k) {
  long long pp = p * k, qq = q * k;
  return pp >= x && qq >= y && qq - y >= pp - x;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      cin >> x >> y >> p >> q;
      if (p == q && x != y) {
        puts("-1");
        continue;
      }
      if (x && p == 0) {
        puts("-1");
        continue;
      }
      if (x * q == y * p) {
        puts("0");
        continue;
      }
      long long l = 1, r = 2e9;
      while (l <= r) {
        long long mid = (l + r) >> 1;
        if (ok(mid)) {
          r = mid - 1;
        } else {
          l = mid + 1;
        }
      }
      cout << q * l - y << '\n';
    }
  }
}

```