#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5, M = 1e12;
map<long long, int> mp;
map<long long, int>::iterator it;
vector<vector<pair<int, int> > > g;
queue<int> qu;
vector<int> v;
long long x, t, y, n = 0, m = 0, u = 0, d = 1, p, z = 1e12, mx = 1e6, mn,
                   sum = 0;
long long l = 0, r = 1e9, mid;
string s;
bool calc(long long w) {
  long long s1 = w * n * d;
  long long s2 = w * m * d;
  if (s2 - y >= s1 - x && s1 - x >= 0 && s2 - y >= 0) {
    u = s2 - y;
    return true;
  } else {
    l = mid + 1;
    return false;
  }
}
int main() {
  cin >> t;
  while (t--) {
    cin >> x >> y >> n >> m;
    l = 0;
    r = 1e9;
    z = 1e18;
    while (l <= r) {
      mid = (l + r) / 2;
      u = 1e18;
      if (calc(mid)) {
        z = min(z, u);
        r = mid - 1;
      }
    }
    if (z == 1e18)
      cout << -1 << endl;
    else
      cout << z << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 1e6 + 5, M = 1e12;
map<long long, int> mp;
map<long long, int>::iterator it;
vector<vector<pair<int, int> > > g;
queue<int> qu;
vector<int> v;
long long x, t, y, n = 0, m = 0, u = 0, d = 1, p, z = 1e12, mx = 1e6, mn,
                   sum = 0;
long long l = 0, r = 1e9, mid;
string s;
bool calc(long long w) {
  long long s1 = w * n * d;
  long long s2 = w * m * d;
  if (s2 - y >= s1 - x && s1 - x >= 0 && s2 - y >= 0) {
    u = s2 - y;
    return true;
  } else {
    l = mid + 1;
    return false;
  }
}
int main() {
  cin >> t;
  while (t--) {
    cin >> x >> y >> n >> m;
    l = 0;
    r = 1e9;
    z = 1e18;
    while (l <= r) {
      mid = (l + r) / 2;
      u = 1e18;
      if (calc(mid)) {
        z = min(z, u);
        r = mid - 1;
      }
    }
    if (z == 1e18)
      cout << -1 << endl;
    else
      cout << z << endl;
  }
  return 0;
}

```