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;
pair<double, double> p[200200];
double dp[200200];
double sy, a, b;
int n, q;
double first, second;
double arr[200200];
double calc(double s) {
  int id = lower_bound(arr + 1, arr + n + 1, s) - arr - 1;
  if (id == 0) return 0;
  double ans = dp[id - 1];
  if (p[id].second > s)
    ans += s - p[id].first;
  else
    ans += p[id].second - p[id].first;
  return ans;
}
int main() {
  scanf("%lf %lf %lf %d", &sy, &a, &b, &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf %lf", &p[i].first, &p[i].second);
    dp[i] = dp[i - 1] + p[i].second - p[i].first;
    arr[i] = p[i].first;
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%lf %lf", &first, &second);
    double tmp = second / (second - sy);
    double pos = first + (b - first) * tmp;
    double ans = calc(pos);
    pos = first + (a - first) * tmp;
    ans -= calc(pos);
    printf("%.10lf\n", ans / tmp);
  }
}
