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 ldouble = double;
ldouble intersect(ldouble x1, ldouble y1, ldouble x2, ldouble y2) {
  ldouble r = y1 / (y1 - y2);
  return x1 + (x2 - x1) * r;
}
void solve() {
  ldouble sy, sl, sr;
  std::cin >> sy >> sl >> sr;
  ldouble t = sr - sl;
  int n;
  std::cin >> n;
  std::vector<std::pair<ldouble, ldouble>> segs(n);
  std::vector<ldouble> ls(n);
  for (int i = 0; i < n; ++i) {
    int l, r;
    std::cin >> l >> r;
    segs[i].first = ls[i] = l;
    segs[i].second = r;
  }
  std::vector<ldouble> sum(n + 1, 0);
  for (int i = 0; i < n; ++i) {
    sum[i + 1] = sum[i] + segs[i].second - segs[i].first;
  }
  auto shade = [&](ldouble x) -> ldouble {
    int i = std::lower_bound(ls.begin(), ls.end(), x) - ls.begin();
    ldouble ret = 0;
    if (i > 0 && x < segs[i - 1].second) {
      ret += x - segs[i - 1].first;
      --i;
    }
    ret += sum[i];
    return ret;
  };
  int q;
  std::cin >> q;
  while (q--) {
    ldouble x, y;
    {
      int xx, yy;
      std::cin >> xx >> yy;
      x = xx, y = yy;
    }
    ldouble l = intersect(x, y, sl, sy), r = intersect(x, y, sr, sy);
    std::cout << (shade(r) - shade(l)) / (r - l) * t << "\n";
  }
}
int main() {
  std::cin.tie(nullptr);
  std::ios::sync_with_stdio(false);
  std::cout << std::fixed << std::setprecision(10);
  solve();
  return 0;
}
