#include <bits/stdc++.h>
using namespace std;
struct bigNum {
  vector<int> val;
  bigNum() { val.push_back(0); }
  void print() {
    printf("%d", val[val.size() - 1]);
    for (int i = val.size() - 2; i >= 0; i--) printf("%05d", val[i]);
  }
  bigNum& operator+=(const bigNum& b) {
    int n = max(val.size(), b.val.size());
    int res = 0;
    for (int i = 0; i < n; i++) {
      if (i < b.val.size()) res += b.val[i];
      if (i < this->val.size())
        res += this->val[i];
      else
        this->val.push_back(0);
      this->val[i] = res % (100000);
      res /= 100000;
    }
    if (res != 0) this->val.push_back(res);
    return *this;
  }
  bigNum& operator=(const string& x) {
    this->val.clear();
    int n = x.length();
    int res = 0;
    for (int i = n - 1; i >= 0; i--) {
      res = res * 10 + (x[i] - '0');
      if (res > 100000) {
        this->val.push_back(res % 100000);
        res /= 100000;
      }
    }
    if (res > 0) this->val.push_back(res % 100000);
    return *this;
  }
  void divide2() {
    bool inc = false;
    int n = this->val.size();
    for (int i = n - 1; i >= 0; i++) {
      int num = this->val[i] + (inc ? 10 : 0);
      this->val[i] = num / 2;
      inc = (num % 2 == 1);
    }
    if (this->val[n - 1] == 0) this->val.erase(this->val.end() - 1);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    if (p == 0) {
      cout << (x == 0 ? "0\n" : "-1\n");
      continue;
    }
    if (p == q) {
      cout << (x == y ? "0\n" : "-1\n");
      continue;
    }
    long long low = 1, high = 9999999999;
    bool poss = false;
    while (low < high) {
      long long mid = (low + high) >> 1;
      if (q * mid >= y && p * mid >= x && q * mid - y >= p * mid - x) {
        high = mid;
        poss = true;
      } else
        low = mid + 1;
    }
    if (poss)
      cout << q * low - y << endl;
    else
      cout << "-1\n";
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
struct bigNum {
  vector<int> val;
  bigNum() { val.push_back(0); }
  void print() {
    printf("%d", val[val.size() - 1]);
    for (int i = val.size() - 2; i >= 0; i--) printf("%05d", val[i]);
  }
  bigNum& operator+=(const bigNum& b) {
    int n = max(val.size(), b.val.size());
    int res = 0;
    for (int i = 0; i < n; i++) {
      if (i < b.val.size()) res += b.val[i];
      if (i < this->val.size())
        res += this->val[i];
      else
        this->val.push_back(0);
      this->val[i] = res % (100000);
      res /= 100000;
    }
    if (res != 0) this->val.push_back(res);
    return *this;
  }
  bigNum& operator=(const string& x) {
    this->val.clear();
    int n = x.length();
    int res = 0;
    for (int i = n - 1; i >= 0; i--) {
      res = res * 10 + (x[i] - '0');
      if (res > 100000) {
        this->val.push_back(res % 100000);
        res /= 100000;
      }
    }
    if (res > 0) this->val.push_back(res % 100000);
    return *this;
  }
  void divide2() {
    bool inc = false;
    int n = this->val.size();
    for (int i = n - 1; i >= 0; i++) {
      int num = this->val[i] + (inc ? 10 : 0);
      this->val[i] = num / 2;
      inc = (num % 2 == 1);
    }
    if (this->val[n - 1] == 0) this->val.erase(this->val.end() - 1);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long x, y, p, q;
    cin >> x >> y >> p >> q;
    if (p == 0) {
      cout << (x == 0 ? "0\n" : "-1\n");
      continue;
    }
    if (p == q) {
      cout << (x == y ? "0\n" : "-1\n");
      continue;
    }
    long long low = 1, high = 9999999999;
    bool poss = false;
    while (low < high) {
      long long mid = (low + high) >> 1;
      if (q * mid >= y && p * mid >= x && q * mid - y >= p * mid - x) {
        high = mid;
        poss = true;
      } else
        low = mid + 1;
    }
    if (poss)
      cout << q * low - y << endl;
    else
      cout << "-1\n";
  }
  return 0;
}

```