#include <bits/stdc++.h>
using namespace std;
const long long int MAX = 1e6 + 5;
const int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int mx = 1e16;
bool comp(const int a, const int b) { return a > b; }
const double PI = 2 * acos(0.0);
long long int fac[N];
void pre() {
  fac[0] = 1LL;
  for (long long int i = 1; i < N; i++) {
    fac[i] = 1LL * i * fac[i - 1];
    if (fac[i] >= mod) fac[i] %= mod;
  }
}
long long int power(long long int a, long long int b) {
  long long int x = 1;
  while (b) {
    if (b & 1) {
      x *= a;
      if (x >= mod) x %= mod;
    }
    a *= a;
    if (a >= mod) a %= mod;
    b >>= 1;
  }
  return x;
}
long long int inv(long long int k) { return power(k, mod - 2); }
string strngtoint(int n) {
  ostringstream str1;
  str1 << n;
  string ans = str1.str();
  return ans;
}
bool fun(long long int z, long long int x, long long int y, long long int p,
         long long int q) {
  if (p * z >= x && q * z - p * z >= y - x)
    return true;
  else
    return false;
}
int main() {
  long long t;
  scanf("%lld\n", &t);
  while (t--) {
    long long int x, y, p, q;
    scanf("%lld %lld %lld %lld", &x, &y, &p, &q);
    long long int low = -1, high = 1e9, mid;
    if (!fun(high, x, y, p, q)) {
      cout << -1 << endl;
      continue;
    }
    while (high - low > 1) {
      mid = (low + high) / 2;
      if (fun(mid, x, y, p, q)) {
        high = mid;
      } else {
        low = mid;
      }
    }
    cout << high * q - y << endl;
  }
}

### Prompt

Generate 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 long long int MAX = 1e6 + 5;
const int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int mx = 1e16;
bool comp(const int a, const int b) { return a > b; }
const double PI = 2 * acos(0.0);
long long int fac[N];
void pre() {
  fac[0] = 1LL;
  for (long long int i = 1; i < N; i++) {
    fac[i] = 1LL * i * fac[i - 1];
    if (fac[i] >= mod) fac[i] %= mod;
  }
}
long long int power(long long int a, long long int b) {
  long long int x = 1;
  while (b) {
    if (b & 1) {
      x *= a;
      if (x >= mod) x %= mod;
    }
    a *= a;
    if (a >= mod) a %= mod;
    b >>= 1;
  }
  return x;
}
long long int inv(long long int k) { return power(k, mod - 2); }
string strngtoint(int n) {
  ostringstream str1;
  str1 << n;
  string ans = str1.str();
  return ans;
}
bool fun(long long int z, long long int x, long long int y, long long int p,
         long long int q) {
  if (p * z >= x && q * z - p * z >= y - x)
    return true;
  else
    return false;
}
int main() {
  long long t;
  scanf("%lld\n", &t);
  while (t--) {
    long long int x, y, p, q;
    scanf("%lld %lld %lld %lld", &x, &y, &p, &q);
    long long int low = -1, high = 1e9, mid;
    if (!fun(high, x, y, p, q)) {
      cout << -1 << endl;
      continue;
    }
    while (high - low > 1) {
      mid = (low + high) / 2;
      if (fun(mid, x, y, p, q)) {
        high = mid;
      } else {
        low = mid;
      }
    }
    cout << high * q - y << endl;
  }
}

```