There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <typename A, typename B = A>
using pr = pair<A, B>;
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  cout << fixed << setprecision(10);
  int sy, a, b;
  cin >> sy >> a >> b;
  sy = -sy;
  int n;
  cin >> n;
  vector<pr<int>> segs(n);
  for (auto &[x, y] : segs) cin >> y >> x;
  sort((segs).begin(), (segs).end());
  vector<ll> pref(n + 1);
  for (int i = 0; i < n; i++)
    pref[i + 1] = pref[i] + segs[i].first - segs[i].second;
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    auto get = [&](int k) { return (k - x) * double(y) / (sy + y); };
    double x1 = x + get(a);
    double x2 = x + get(b);
    auto find = [&](double x) {
      return int(upper_bound((segs).begin(), (segs).end(),
                             pr<int>{ceil(x + 1e-6), -1}) -
                 segs.begin());
    };
    int i1 = find(x1);
    int i2 = find(x2);
    double sum = 0;
    if (i2 != i1) {
      sum += pref[i2] - pref[i1 + 1];
      if (i1 != -1)
        sum +=
            min<double>(segs[i1].first - segs[i1].second, segs[i1].first - x1);
      if (i2 != n) sum += max<double>(0, x2 - segs[i2].second);
    } else {
      int i = i2;
      if (i != -1 && i != n)
        sum += max<double>(0, min<double>(segs[i].first, x2) -
                                  max<double>(segs[i].second, x1));
    }
    cout << sum * (sy + y) / y << '\n';
  }
}
