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 = 200050;
const double INF = 1e10;
struct point {
  double x, y;
  point(double x0 = 0, double y0 = 0) : x(x0), y(y0) {}
};
double l[maxn], r[maxn], sum[maxn];
point p[maxn];
void solve() {
  double x1, x2, y;
  scanf("%lf %lf %lf", &y, &x1, &x2);
  int n, q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lf %lf", &l[i], &r[i]);
  l[0] = 0;
  r[0] = 0;
  l[n + 1] = INF;
  r[n + 1] = INF;
  for (int i = 1; i <= n + 1; i++) sum[i] = sum[i - 1] + r[i] - l[i];
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%lf %lf", &p[i].x, &p[i].y);
    double res = 0;
    double _l = p[i].x + (x1 - p[i].x) * p[i].y / (p[i].y - y);
    double _r = p[i].x + (x2 - p[i].x) * p[i].y / (p[i].y - y);
    int a = lower_bound(l, l + n + 2, _l) - l;
    int b = lower_bound(r, r + n + 2, _r) - r;
    if (_l >= l[a - 1] && _l <= r[a - 1]) {
      res -= sum[a - 1];
      res += r[a - 1] - _l;
    } else if (_l > r[a - 1]) {
      res -= sum[a - 1];
    }
    if (_r < l[b]) {
      res += sum[b - 1];
    } else if (_r >= l[b] && _r <= r[b]) {
      res += sum[b];
      res -= r[b] - _r;
    }
    printf("%.15f\n", res * (x2 - x1) / (_r - _l));
  }
}
int main() {
  solve();
  return 0;
}
