#include <bits/stdc++.h>
using namespace std;
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long nCr(long long n, long long r) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = min(i, r); j > 0; j--) C[j] = C[j] + C[j - 1];
  return C[r];
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n == 2) return true;
  if (n % 2 == 0) return false;
  long long m = sqrt(n);
  for (long long i = 3; i <= m; i += 2)
    if (n % i == 0) return false;
  return true;
}
long long M(long long a) { return (a % (1000000007LL)); }
bool isPowerOfTwo(long long x) { return x && (!(x & (x - 1))); }
int x, y, p, q;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    if (x == 0 && p == 0) {
      cout << 0 << "\n";
      continue;
    } else if (x == y && p == q) {
      cout << 0 << "\n";
      continue;
    } else if (p == q && x != y) {
      cout << -1 << "\n";
      continue;
    } else if (p == 0 && x != 0) {
      cout << -1 << "\n";
      continue;
    }
    int l = 1, r = 1e9 + 1;
    int mid;
    int f = (x + p - 1) / p;
    int s = (y - x + q - p - 1) / (q - p);
    while (r > l) {
      mid = (l + r) >> 1;
      if (mid >= f && mid >= s) {
        r = mid;
      } else {
        l = mid + 1;
      }
      if (((l + r) >> 1) == mid) {
        break;
      }
    }
    if (mid >= f && mid >= s) {
      cout << (1LL * q * 1LL * mid - 1LL * y * 1LL) << "\n";
    } else if (l >= f && l >= s) {
      cout << (1LL * q * 1LL * l - 1LL * y * 1LL) << "\n";
    } else if (r >= f && r >= s) {
      cout << (1LL * q * 1LL * r - 1LL * y * 1LL) << "\n";
    } else
      cout << -1 << "\n";
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long nCr(long long n, long long r) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = min(i, r); j > 0; j--) C[j] = C[j] + C[j - 1];
  return C[r];
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n == 2) return true;
  if (n % 2 == 0) return false;
  long long m = sqrt(n);
  for (long long i = 3; i <= m; i += 2)
    if (n % i == 0) return false;
  return true;
}
long long M(long long a) { return (a % (1000000007LL)); }
bool isPowerOfTwo(long long x) { return x && (!(x & (x - 1))); }
int x, y, p, q;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> x >> y >> p >> q;
    if (x == 0 && p == 0) {
      cout << 0 << "\n";
      continue;
    } else if (x == y && p == q) {
      cout << 0 << "\n";
      continue;
    } else if (p == q && x != y) {
      cout << -1 << "\n";
      continue;
    } else if (p == 0 && x != 0) {
      cout << -1 << "\n";
      continue;
    }
    int l = 1, r = 1e9 + 1;
    int mid;
    int f = (x + p - 1) / p;
    int s = (y - x + q - p - 1) / (q - p);
    while (r > l) {
      mid = (l + r) >> 1;
      if (mid >= f && mid >= s) {
        r = mid;
      } else {
        l = mid + 1;
      }
      if (((l + r) >> 1) == mid) {
        break;
      }
    }
    if (mid >= f && mid >= s) {
      cout << (1LL * q * 1LL * mid - 1LL * y * 1LL) << "\n";
    } else if (l >= f && l >= s) {
      cout << (1LL * q * 1LL * l - 1LL * y * 1LL) << "\n";
    } else if (r >= f && r >= s) {
      cout << (1LL * q * 1LL * r - 1LL * y * 1LL) << "\n";
    } else
      cout << -1 << "\n";
  }
  return 0;
}

```