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 y, a, b;
int n1, n2;
const double eps = 1e-9;
pair<int, int> engel[200007];
map<int, int> pre;
double kesis(double a1, double a2, double b1, double b2) {
  if (a1 >= b2 || b1 >= a2) return 0;
  if ((a1 <= b1 && a2 >= b2) || (b1 <= a1 && b2 >= a2))
    return min(a2 - a1, b2 - b1);
  if (a1 <= b1) return a2 - b1;
  return b2 - a1;
}
double solve(double xx, double yy) {
  double l = xx - ((xx - a) / (yy - y) * yy),
         r = xx - ((xx - b) / (yy - y) * yy);
  int ll = 0, rr = n1 - 1, m;
  int la = -1e9 - 100;
  while (ll <= rr) {
    m = (ll + rr) / 2;
    if (engel[m].first <= l) {
      ll = m + 1;
      la = max(la, m);
    } else
      rr = m - 1;
  }
  int k1 = max(la, -1);
  ll = 0;
  rr = n1 - 1;
  la = 1e9 + 10;
  while (ll <= rr) {
    m = (ll + rr) / 2;
    if (engel[m].second >= r) {
      rr = m - 1;
      la = min(la, m);
    } else
      ll = m + 1;
  }
  int k2 = la;
  k2 = min(k2, n1);
  double ans = 0;
  if (k1 != k2) ans = pre[k2 - 1] - pre[k1];
  if (k1 != -1) ans += kesis(l, r, engel[k1].first, engel[k1].second);
  if (k2 != n1) ans += kesis(l, r, engel[k2].first, engel[k2].second);
  if (k1 == k2 && k1 != -1 && k1 != n1)
    ans -= kesis(l, r, engel[k2].first, engel[k2].second);
  return ans * (b - a) / (r - l);
}
int main() {
  scanf("%lf %lf %lf", &y, &a, &b);
  scanf("%d ", &n1);
  int i;
  int l = 0;
  for (i = 0; i < n1; i++) {
    scanf("%d %d", &engel[i].first, &engel[i].second);
    l += engel[i].second - engel[i].first;
    pre[i] = l;
  }
  scanf("%d ", &n2);
  double x, y;
  for (i = 0; i < n2; i++) {
    scanf("%lf %lf", &x, &y);
    printf("%.12lf\n", solve(x, y));
  }
  return 0;
}
