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;
int sy, a, b, n, q, l[200003], r[200003], cnt[200003];
double YY(double f, double t) {
  int ll = 0, rr = n;
  while (ll != rr) {
    int mid = (ll + rr) / 2;
    if (l[mid] <= f)
      ll = mid + 1;
    else
      rr = mid;
  }
  int left = ll, right;
  ll = 0, rr = n;
  while (ll != rr) {
    int mid = (ll + rr) / 2;
    if (l[mid] <= t)
      ll = mid + 1;
    else
      rr = mid;
  }
  right = ll;
  double k = 0;
  return cnt[right] - cnt[left] + (left ? max(k, r[left - 1] - f) : 0) -
         (right ? max(k, r[right - 1] - t) : 0);
}
int main() {
  scanf("%d%d%d%d", &sy, &a, &b, &n);
  for (int i = 0; i < n; i++)
    scanf("%d%d", l + i, r + i), cnt[i + 1] = cnt[i] + r[i] - l[i];
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    double lth = y * 1.0 / (y - sy);
    cout << fixed << setprecision(10)
         << YY(a * lth + x * 1.0 / (sy - y) * sy,
               b * lth + x * 1.0 / (sy - y) * sy) /
                lth
         << endl;
  }
}
