#include <bits/stdc++.h>
using namespace std;
pair<int, int> g[32000];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int a, b, p, q;
    cin >> a >> b >> p >> q;
    int n = sqrt(b - a + 1);
    for (int i = 0; i < n; i++)
      g[i + 1] = make_pair(2ll * p * (a + i) % (2 * q), i);
    sort(g + 1, g + n + 1);
    int m = 1;
    for (int i = 2; i <= n; i++)
      if (g[i].first != g[i - 1].first) g[++m] = g[i];
    g[0] = g[m], g[0].first -= 2 * q;
    g[m + 1] = g[1], g[m + 1].first += 2 * q;
    pair<int, int> s(2 * q, 0);
    for (int i = 0; i < n; i++) {
      int fix = ((long long)(2 * q - 2ll) * p % (2 * q) * i * n + q) % (2 * q);
      int x = lower_bound(g + 1, g + m + 1, make_pair(fix, 0)) - g - 1;
      s = min(s, make_pair(abs(fix - g[x].first), g[x].second + i * n + a));
      s = min(
          s, make_pair(abs(fix - g[x + 1].first), g[x + 1].second + i * n + a));
    }
    for (int i = a + n * n; i <= b; i++)
      s = min(s, make_pair((int)abs(q - 2ll * i * p % (2 * q)), i));
    cout << s.second << endl;
  }
}

### Prompt

Generate a CPP solution to the following problem:

You have given integers a, b, p, and q. Let f(x) = abs(sin(p/q π x)).

Find minimum possible integer x that maximizes f(x) where a ≤ x ≤ b.

Input

Each test contains multiple test cases. 

The first line contains the number of test cases t (1 ≤ t ≤ 100) — the number of test cases.

The first line of each test case contains four integers a, b, p, and q (0 ≤ a ≤ b ≤ 10^{9}, 1 ≤ p, q ≤ 10^{9}).

Output

Print the minimum possible integer x for each test cases, separated by newline.

Example

Input


2
0 3 1 3
17 86 389 995


Output


1
55

Note

In the first test case, f(0) = 0, f(1) = f(2) ≈ 0.866, f(3) = 0.

In the second test case, f(55) ≈ 0.999969, which is the largest among all possible values.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
pair<int, int> g[32000];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int a, b, p, q;
    cin >> a >> b >> p >> q;
    int n = sqrt(b - a + 1);
    for (int i = 0; i < n; i++)
      g[i + 1] = make_pair(2ll * p * (a + i) % (2 * q), i);
    sort(g + 1, g + n + 1);
    int m = 1;
    for (int i = 2; i <= n; i++)
      if (g[i].first != g[i - 1].first) g[++m] = g[i];
    g[0] = g[m], g[0].first -= 2 * q;
    g[m + 1] = g[1], g[m + 1].first += 2 * q;
    pair<int, int> s(2 * q, 0);
    for (int i = 0; i < n; i++) {
      int fix = ((long long)(2 * q - 2ll) * p % (2 * q) * i * n + q) % (2 * q);
      int x = lower_bound(g + 1, g + m + 1, make_pair(fix, 0)) - g - 1;
      s = min(s, make_pair(abs(fix - g[x].first), g[x].second + i * n + a));
      s = min(
          s, make_pair(abs(fix - g[x + 1].first), g[x + 1].second + i * n + a));
    }
    for (int i = a + n * n; i <= b; i++)
      s = min(s, make_pair((int)abs(q - 2ll * i * p % (2 * q)), i));
    cout << s.second << endl;
  }
}

```