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>
#pragma GCC optimize("O3")
using namespace std;
double l[200005], r[200005], a[200005];
int main() {
  double sy, sl, sr;
  int n, q;
  scanf("%lf%lf%lf%d", &sy, &sl, &sr, &n);
  for (int i = 1; i <= n; ++i)
    scanf("%lf%lf", &l[i], &r[i]), a[i] = a[i - 1] + r[i] - l[i];
  scanf("%d", &q);
  while (q--) {
    double x, y;
    scanf("%lf%lf", &x, &y);
    double xl = sl + (x - sl) * (-sy) / (y - sy),
           xr = sr + (x - sr) * (-sy) / (y - sy);
    int l2 = upper_bound(l + 1, l + n + 1, xl + 1e-6) - l - 1,
        r2 = upper_bound(l + 1, l + n + 1, xr + 1e-6) - l - 1;
    double l3 = l2 > 0 ? (a[l2 - 1] + min(r[l2], xl) - l[l2]) : 0,
           r3 = r2 > 0 ? (a[r2 - 1] + min(r[r2], xr) - l[r2]) : 0;
    printf("%.10f\n", (r3 - l3) * (y - sy) / y);
  }
}
