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;
double SY, A, B;
int num_pts = 0;
int N, Q;
double ans[200000];
double ox[500000];
double val[500000];
double bin_exact(double T) {
  int l = 0, r = num_pts;
  while (l < r) {
    int mid = (l + r) / 2;
    double x = ox[mid];
    if (x <= T)
      l = mid + 1;
    else if (x > T)
      r = mid;
  }
  return val[l - 1];
}
pair<double, double> bin_down(double T) {
  int l = 0, r = num_pts;
  while (l < r) {
    int mid = (l + r) / 2;
    double x = ox[mid];
    if (x <= T)
      l = mid + 1;
    else
      r = mid;
  }
  return make_pair(ox[l - 1], val[l - 1]);
}
pair<double, double> bin_up(double T) {
  int l = 0, r = num_pts;
  while (l < r) {
    int mid = (l + r) / 2;
    double x = ox[mid];
    if (x <= T)
      l = mid + 1;
    else
      r = mid;
  }
  return make_pair(ox[l], val[l]);
}
double func(double x, double y) {
  double _l, _r;
  if (x == A && false) {
    _l = bin_exact(x);
  } else {
    double t = A - x;
    t *= y / (y - SY);
    t += x;
    pair<double, double> LL = bin_down(t);
    pair<double, double> RR = bin_up(t);
    double tl = LL.first;
    double tr = RR.first;
    _l = (t - tl) / (tr - tl) * (RR.second - LL.second) + LL.second;
  }
  if (x == B && false) {
    _r = bin_exact(x);
  } else {
    double t = B - x;
    t *= y / (y - SY);
    t += x;
    pair<double, double> LL = bin_down(t);
    pair<double, double> RR = bin_up(t);
    double tl = LL.first;
    double tr = RR.first;
    _r = (t - tl) / (tr - tl) * (RR.second - LL.second) + LL.second;
  }
  return (_r - _l) * ((y - SY) / y);
}
int main(int argc, char **argv) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int _SY, _A, _B;
  cin >> _SY >> _A >> _B;
  SY = _SY, A = _A, B = _B;
  cin >> N;
  ox[num_pts] = 0.0;
  val[num_pts] = 0.0;
  num_pts++;
  double tmp = 0;
  for (int i = 0; i < N; i++) {
    int a, b;
    cin >> a >> b;
    ox[num_pts] = a;
    val[num_pts] = tmp;
    num_pts++;
    tmp += b - a;
    ox[num_pts] = b;
    val[num_pts] = tmp;
    num_pts++;
  }
  ox[num_pts] = 2000000000.0;
  val[num_pts] = tmp;
  num_pts++;
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int x, y;
    cin >> x >> y;
    ans[i] = func(x, y);
  }
  cout << fixed << setprecision(7);
  for (int i = 0; i < Q; i++) {
    cout << ans[i] << '\n';
  }
  return 0;
}
