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.

<image>

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;
int main() {
  int a, b, sy;
  scanf("%d%d%d", &(sy), &(a), &(b));
  sy *= -1;
  set<pair<double, pair<int, int>>> m;
  int n, q;
  scanf("%d", &(n));
  int mr;
  for (int i = 0; i < (n); ++i) {
    int l, r;
    scanf("%d%d", &(l), &(r));
    m.insert({double(l), {1, 0}});
    m.insert({double(r), {-1, 0}});
    mr = r;
  }
  scanf("%d", &(q));
  for (int i = 0; i < (q); ++i) {
    int x, y;
    scanf("%d%d", &(x), &(y));
    double s, e;
    s = (double)a + (double(x - a) * sy) / (double(sy) + y);
    e = (double)b + (double(x - b) * sy) / (double(sy) + y);
    m.insert({s, {0, i + 1}});
    m.insert({e, {0, -i - 1}});
  }
  int hidden = 0;
  vector<pair<double, double>> rr(q);
  vector<double> rrr(q);
  double previous = 0;
  double disthidden = 0;
  double distot = 0;
  for (auto i : m) {
    if (hidden) {
      disthidden += (i.first - previous);
    }
    previous = i.first;
    if (!i.second.first) {
      if (i.second.second > 0) {
        rr[i.second.second - 1] = {disthidden, i.first};
      } else {
        int ix = -(i.second.second + 1);
        rrr[ix] = (double)(b - a) * (disthidden - rr[ix].first) /
                  (i.first - rr[ix].second);
      }
    }
    hidden += i.second.first;
  }
  for (int i = 0; i < (q); ++i) {
    printf("%.8f\n", rrr[i]);
  }
}
