#include <bits/stdc++.h>
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (a ? gcd(b % a, a) : b);
}
long long int modPow(long long int a, long long int b, long long int MOD) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % MOD;
    }
    b /= 2;
    y = (y * y) % MOD;
  }
  return x;
}
long long int modInverse(long long int a, long long int p) {
  return modPow(a, p - 2, p);
}
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int tc, x, y, p, q;
  cin >> tc;
  while (tc--) {
    cin >> x >> y >> p >> q;
    double a = (double)x / y;
    double b = (double)p / q;
    if (a == b) {
      cout << "0\n";
      continue;
    }
    if (b == 0.0) {
      if (x != 0)
        cout << "-1\n";
      else
        cout << "0\n";
    } else if (b == 1.0) {
      if (x == y)
        cout << "0\n";
      else
        cout << "-1\n";
    } else {
      long long int ans = -1;
      long long int hi = 1e10;
      long long int lo = 0;
      long long int mid;
      while (lo <= hi) {
        mid = (lo + hi) / 2;
        long long int db = mid * q - y;
        long long int da = mid * p - x;
        if (da >= 0 && db >= 0 && da <= db) {
          ans = db;
          hi = mid - 1;
        } else
          lo = mid + 1;
      }
      cout << ans << "\n";
    }
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
long long int gcd(long long int a, long long int b) {
  return (a ? gcd(b % a, a) : b);
}
long long int modPow(long long int a, long long int b, long long int MOD) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % MOD;
    }
    b /= 2;
    y = (y * y) % MOD;
  }
  return x;
}
long long int modInverse(long long int a, long long int p) {
  return modPow(a, p - 2, p);
}
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int tc, x, y, p, q;
  cin >> tc;
  while (tc--) {
    cin >> x >> y >> p >> q;
    double a = (double)x / y;
    double b = (double)p / q;
    if (a == b) {
      cout << "0\n";
      continue;
    }
    if (b == 0.0) {
      if (x != 0)
        cout << "-1\n";
      else
        cout << "0\n";
    } else if (b == 1.0) {
      if (x == y)
        cout << "0\n";
      else
        cout << "-1\n";
    } else {
      long long int ans = -1;
      long long int hi = 1e10;
      long long int lo = 0;
      long long int mid;
      while (lo <= hi) {
        mid = (lo + hi) / 2;
        long long int db = mid * q - y;
        long long int da = mid * p - x;
        if (da >= 0 && db >= 0 && da <= db) {
          ans = db;
          hi = mid - 1;
        } else
          lo = mid + 1;
      }
      cout << ans << "\n";
    }
  }
}

```