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;
const int maxn = 2e5 + 5;
double sy, sa, sb;
int n;
double l[maxn], r[maxn];
double sum[maxn];
double calx(double x, double y, double ssa) {
  double tmp = y * ssa - x * sy;
  double tmp2 = y - sy;
  return 1.0 * tmp / tmp2;
}
int main() {
  scanf("%lf%lf%lf", &sy, &sa, &sb);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf%lf", &l[i], &r[i]);
  }
  for (int i = 1; i <= n + 1; i++) {
    sum[i] = sum[i - 1] + (r[i] - l[i]);
  }
  int q;
  scanf("%d", &q);
  double x1, x2;
  while (q--) {
    double x, y;
    scanf("%lf%lf", &x, &y);
    double x1 = calx(x, y, sa);
    double x2 = calx(x, y, sb);
    int pos1 = lower_bound(r + 1, r + 1 + n, (ceil(x1)) - 0.1) - r;
    int pos2 = lower_bound(r + 1, r + 1 + n, (ceil(x2)) - 0.1) - r;
    int st = pos1 + 1;
    int en = pos2 - 1;
    double ans = 0;
    if (en >= st) ans += sum[en] - sum[st - 1];
    if (x1 >= l[pos1])
      ans += r[pos1] - x1;
    else if (pos1 != pos2)
      ans += r[pos1] - l[pos1];
    else if (x2 > l[pos1])
      ans = x2 - l[pos1];
    else
      ans = 0;
    if (pos2 != n + 1 && x2 >= l[pos2])
      if (pos1 != pos2) ans += x2 - l[pos2];
    if (pos1 == pos2 && l[pos1] <= x1 && x2 <= r[pos1])
      ans = min(r[pos1], x2) - max(x1, l[pos1]);
    ans = 1.0 * ans * (sb - sa) / (x2 - x1);
    printf("%.10f\n", max(0.0, ans));
  }
  return 0;
}
