#include <bits/stdc++.h>
using namespace std;
unsigned long long mod = 1e9 + 7;
long long MOD = 998244353;
unsigned long long ncr(long long n, long long k) {
  if (k > n) return 1ll * 0;
  unsigned long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
    res = res;
  }
  return res;
}
bool prime(unsigned long long num) {
  bool flag = true;
  for (unsigned long long i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      flag = false;
      break;
    }
  }
  return flag;
}
long long ipow(long long base, long long exp) {
  if (exp == 0) return 1ll * 1;
  long long result = 1;
  for (;;) {
    if (exp & 1) {
      result *= base;
      result = result % MOD;
    }
    exp >>= 1;
    if (!exp) break;
    base *= base;
    base = base % MOD;
  }
  return result % MOD;
}
long long abs(long long a, long long b) {
  if ((a - b) > 0)
    return (a - b);
  else
    return (b - a);
}
long long add(long long a, long long b) {
  a += b;
  if (a >= MOD)
    a -= MOD;
  else if (a < 0)
    a += MOD;
  return a;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long der(long long k) {
  long long res = 0;
  long long fact = 1;
  for (long long i = 2; i <= k; i++) fact *= i;
  long long s = 1;
  long long j = 1;
  for (long long i = 2; i <= k; i++) {
    j = j * i;
    long long first = (fact / j) * s;
    res += first;
    s = s * (-1);
  }
  return res;
}
int len;
string p;
bool check(string s, vector<int> v, int val) {
  vector<int> rem;
  for (int i = 0; i < val; i++) rem.push_back(v[i]);
  sort((rem).begin(), (rem).end());
  int ind = 0;
  int nind = 0;
  for (int i = 0; i < len && nind != p.size(); i++) {
    if (ind != rem.size()) {
      if (i == rem[ind] - 1) {
        ind++;
        continue;
      }
    }
    if (p[nind] == s[i]) {
      nind++;
    }
  }
  return nind == p.size();
}
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;
    long long ans = -1;
    if (p == q) {
      if (x < y)
        cout << "-1\n";
      else
        cout << "0\n";
      continue;
    }
    if (p == 0) {
      if (x != 0)
        cout << "-1\n";
      else
        cout << "0\n";
      continue;
    }
    long long num = q / y;
    if (q % y != 0) num++;
    if (q >= y) num = 1;
    long long left = num;
    long long rht = LLONG_MAX / q;
    while (left <= rht) {
      long long mid = (left + rht) / 2;
      long long nq = mid * q;
      long long np = mid * p;
      long long ext = nq - y;
      if (x + ext >= np && x <= np) {
        rht = mid - 1;
        ans = nq;
      } else {
        left = mid + 1;
      }
    }
    cout << ans - y << endl;
  }
}

### Prompt

Your challenge is to write 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;
unsigned long long mod = 1e9 + 7;
long long MOD = 998244353;
unsigned long long ncr(long long n, long long k) {
  if (k > n) return 1ll * 0;
  unsigned long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
    res = res;
  }
  return res;
}
bool prime(unsigned long long num) {
  bool flag = true;
  for (unsigned long long i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      flag = false;
      break;
    }
  }
  return flag;
}
long long ipow(long long base, long long exp) {
  if (exp == 0) return 1ll * 1;
  long long result = 1;
  for (;;) {
    if (exp & 1) {
      result *= base;
      result = result % MOD;
    }
    exp >>= 1;
    if (!exp) break;
    base *= base;
    base = base % MOD;
  }
  return result % MOD;
}
long long abs(long long a, long long b) {
  if ((a - b) > 0)
    return (a - b);
  else
    return (b - a);
}
long long add(long long a, long long b) {
  a += b;
  if (a >= MOD)
    a -= MOD;
  else if (a < 0)
    a += MOD;
  return a;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long der(long long k) {
  long long res = 0;
  long long fact = 1;
  for (long long i = 2; i <= k; i++) fact *= i;
  long long s = 1;
  long long j = 1;
  for (long long i = 2; i <= k; i++) {
    j = j * i;
    long long first = (fact / j) * s;
    res += first;
    s = s * (-1);
  }
  return res;
}
int len;
string p;
bool check(string s, vector<int> v, int val) {
  vector<int> rem;
  for (int i = 0; i < val; i++) rem.push_back(v[i]);
  sort((rem).begin(), (rem).end());
  int ind = 0;
  int nind = 0;
  for (int i = 0; i < len && nind != p.size(); i++) {
    if (ind != rem.size()) {
      if (i == rem[ind] - 1) {
        ind++;
        continue;
      }
    }
    if (p[nind] == s[i]) {
      nind++;
    }
  }
  return nind == p.size();
}
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;
    long long ans = -1;
    if (p == q) {
      if (x < y)
        cout << "-1\n";
      else
        cout << "0\n";
      continue;
    }
    if (p == 0) {
      if (x != 0)
        cout << "-1\n";
      else
        cout << "0\n";
      continue;
    }
    long long num = q / y;
    if (q % y != 0) num++;
    if (q >= y) num = 1;
    long long left = num;
    long long rht = LLONG_MAX / q;
    while (left <= rht) {
      long long mid = (left + rht) / 2;
      long long nq = mid * q;
      long long np = mid * p;
      long long ext = nq - y;
      if (x + ext >= np && x <= np) {
        rht = mid - 1;
        ans = nq;
      } else {
        left = mid + 1;
      }
    }
    cout << ans - y << endl;
  }
}

```