#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 3e6;
const long long mod = 1e9 + 7;
const long double PI = acos((long double)-1);
long long pw(long long a, long long b, long long md = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (a * res) % md;
    }
    a = (a * a) % md;
    b >>= 1;
  }
  return (res);
}
int t;
long long x, y, p, q;
bool chk(long long mid) {
  if (p * mid < x) return (0);
  if (mid * (q - p) < y) return (0);
  return (1);
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    if (p == q and x != y) {
      cout << -1 << '\n';
      continue;
    }
    if (x == y and p == q) {
      cout << 0 << '\n';
      continue;
    }
    if (p == 0) {
      if (x == 0) {
        cout << 0 << '\n';
      } else {
        cout << -1 << '\n';
      }
      continue;
    }
    long long l = -1, r = 2e9;
    y -= x;
    while (r - l > 1) {
      long long mid = (l + r) / 2;
      if (chk(mid))
        r = mid;
      else
        l = mid;
    }
    cout << r * q - x - y << '\n';
  }
  return (0);
}

### Prompt

Your task is to create 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 3e6;
const long long mod = 1e9 + 7;
const long double PI = acos((long double)-1);
long long pw(long long a, long long b, long long md = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (a * res) % md;
    }
    a = (a * a) % md;
    b >>= 1;
  }
  return (res);
}
int t;
long long x, y, p, q;
bool chk(long long mid) {
  if (p * mid < x) return (0);
  if (mid * (q - p) < y) return (0);
  return (1);
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    if (p == q and x != y) {
      cout << -1 << '\n';
      continue;
    }
    if (x == y and p == q) {
      cout << 0 << '\n';
      continue;
    }
    if (p == 0) {
      if (x == 0) {
        cout << 0 << '\n';
      } else {
        cout << -1 << '\n';
      }
      continue;
    }
    long long l = -1, r = 2e9;
    y -= x;
    while (r - l > 1) {
      long long mid = (l + r) / 2;
      if (chk(mid))
        r = mid;
      else
        l = mid;
    }
    cout << r * q - x - y << '\n';
  }
  return (0);
}

```