#include <bits/stdc++.h>
using namespace std;
long long fact(long long num) {
  long long x = 1;
  for (long long i = 1; i <= num; ++i) {
    x = x * i;
  }
  return x;
}
void swap(long long *a, long long *b) {
  long long x = *b;
  *b = *a;
  *a = x;
}
int isPrime(long long num) {
  if (num <= 1) return 0;
  for (long long i = 2; i * i <= num; i++)
    if (num % i == 0) return 0;
  return 1;
}
long long binomialCoeff(long long n, long long k) {
  long long C[k + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, k); j > 0; j--)
      C[j] = (C[j] + C[j - 1]) % 1000000007;
  }
  return C[k];
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modFact(long long n, long long p) {
  if (n >= p) return 0;
  long long result = 1;
  for (long long i = 1; i <= n; i++) result = (result * i) % p;
  return result;
}
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
long long decrease(int x, int y, int p, int q) {
  long long low = 1;
  long long high = LLONG_MAX;
  long long res = -1;
  while (low <= high) {
    long long mid = low + (high - low) / 2;
    cout << "d" << low << " " << high << " " << mid << endl;
    if ((long double)x / (y + mid) <= (long double)p / q) {
      res = mid;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  return res;
}
long long increase(int x, int y, int p, int q) {
  long long low = 1;
  long long high = LLONG_MAX;
  long long res = -1;
  while (low <= high) {
    long long mid = low + (high - low) / 2;
    if ((long double)(x + mid) / (y + mid) >= (long double)p / q) {
      res = mid;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
    cout << res << endl;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long maxm = 0, ans = 0, sum = 0, count = 0, minm = 1000000000000;
  long long tc, m, n, k, t, x, y, p, q;
  long long f = 0;
  cin >> tc;
  while (tc--) {
    cin >> x >> y >> p >> q;
    if (p == 0) {
      cout << (x == 0 ? 0 : -1) << endl;
      continue;
    }
    if (p == q) {
      cout << (x == y ? 0 : -1) << endl;
      continue;
    }
    t = max((x + p - 1) / p, ((y - x) + (q - p - 1)) / (q - p));
    cout << q * t - y << endl;
  }
  return 0;
}

### Prompt

Please formulate 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;
long long fact(long long num) {
  long long x = 1;
  for (long long i = 1; i <= num; ++i) {
    x = x * i;
  }
  return x;
}
void swap(long long *a, long long *b) {
  long long x = *b;
  *b = *a;
  *a = x;
}
int isPrime(long long num) {
  if (num <= 1) return 0;
  for (long long i = 2; i * i <= num; i++)
    if (num % i == 0) return 0;
  return 1;
}
long long binomialCoeff(long long n, long long k) {
  long long C[k + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, k); j > 0; j--)
      C[j] = (C[j] + C[j - 1]) % 1000000007;
  }
  return C[k];
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modFact(long long n, long long p) {
  if (n >= p) return 0;
  long long result = 1;
  for (long long i = 1; i <= n; i++) result = (result * i) % p;
  return result;
}
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
long long decrease(int x, int y, int p, int q) {
  long long low = 1;
  long long high = LLONG_MAX;
  long long res = -1;
  while (low <= high) {
    long long mid = low + (high - low) / 2;
    cout << "d" << low << " " << high << " " << mid << endl;
    if ((long double)x / (y + mid) <= (long double)p / q) {
      res = mid;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  return res;
}
long long increase(int x, int y, int p, int q) {
  long long low = 1;
  long long high = LLONG_MAX;
  long long res = -1;
  while (low <= high) {
    long long mid = low + (high - low) / 2;
    if ((long double)(x + mid) / (y + mid) >= (long double)p / q) {
      res = mid;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
    cout << res << endl;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long maxm = 0, ans = 0, sum = 0, count = 0, minm = 1000000000000;
  long long tc, m, n, k, t, x, y, p, q;
  long long f = 0;
  cin >> tc;
  while (tc--) {
    cin >> x >> y >> p >> q;
    if (p == 0) {
      cout << (x == 0 ? 0 : -1) << endl;
      continue;
    }
    if (p == q) {
      cout << (x == y ? 0 : -1) << endl;
      continue;
    }
    t = max((x + p - 1) / p, ((y - x) + (q - p - 1)) / (q - p));
    cout << q * t - y << endl;
  }
  return 0;
}

```