#include <bits/stdc++.h>
const int MAX_LEN = 1010;
using namespace std;
template <typename U, typename V>
string to_string(pair<U, V>);
string to_string(const string& e_) { return "\"" + e_ + "\""; }
string to_string(char e_) { return "\'" + string(1, e_) + "\'"; }
string to_string(bool e_) { return e_ ? "true" : "false"; }
template <typename T>
string to_string(T e_) {
  string s_ = "[ ";
  for (const auto& x_ : e_) s_ += to_string(x_) + " ";
  return s_ + "]";
}
template <typename U, typename V>
string to_string(pair<U, V> e_) {
  return "(" + to_string(e_.first) + ", " + to_string(e_.second) + ")";
}
void dbg_str() { ; }
template <typename U, typename... V>
void dbg_str(U u, V... v) {
  ;
  dbg_str(v...);
}
long long int t, x, y, p, q;
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%lld %lld %lld %lld", &x, &y, &p, &q);
    long long int lt = 0;
    long long int rt = 1e9;
    long long int ans = -1;
    long long int mt;
    while (lt <= rt) {
      ;
      mt = (lt + rt) / 2;
      if (mt * p >= x && mt * (q - p) >= (y - x)) {
        ans = mt;
        rt = mt - 1;
      } else
        lt = mt + 1;
    };
    if (ans == -1)
      printf("-1\n");
    else
      printf("%lld\n", q * ans - y);
  }
}

### 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>
const int MAX_LEN = 1010;
using namespace std;
template <typename U, typename V>
string to_string(pair<U, V>);
string to_string(const string& e_) { return "\"" + e_ + "\""; }
string to_string(char e_) { return "\'" + string(1, e_) + "\'"; }
string to_string(bool e_) { return e_ ? "true" : "false"; }
template <typename T>
string to_string(T e_) {
  string s_ = "[ ";
  for (const auto& x_ : e_) s_ += to_string(x_) + " ";
  return s_ + "]";
}
template <typename U, typename V>
string to_string(pair<U, V> e_) {
  return "(" + to_string(e_.first) + ", " + to_string(e_.second) + ")";
}
void dbg_str() { ; }
template <typename U, typename... V>
void dbg_str(U u, V... v) {
  ;
  dbg_str(v...);
}
long long int t, x, y, p, q;
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%lld %lld %lld %lld", &x, &y, &p, &q);
    long long int lt = 0;
    long long int rt = 1e9;
    long long int ans = -1;
    long long int mt;
    while (lt <= rt) {
      ;
      mt = (lt + rt) / 2;
      if (mt * p >= x && mt * (q - p) >= (y - x)) {
        ans = mt;
        rt = mt - 1;
      } else
        lt = mt + 1;
    };
    if (ans == -1)
      printf("-1\n");
    else
      printf("%lld\n", q * ans - y);
  }
}

```