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.

<image>

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;
template <typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;
const int N = 2e5 + 5;
int n, q;
long long l[N], r[N], len[N], x, y, sy, a, b;
inline double inter(long long x, long long y, long long a, long long sy) {
  return a - (1.0 * (x - a) * sy) / (y - sy);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> sy >> a >> b >> n;
  for (int i = (1); i < (n + 1); i++) {
    cin >> l[i] >> r[i];
    len[i] = (r[i] - l[i]) + len[i - 1];
  }
  cin >> q;
  cout << fixed << setprecision(8);
  while (q--) {
    cin >> x >> y;
    double d1 = inter(x, y, a, sy);
    double d2 = inter(x, y, b, sy);
    int i = lower_bound(r + 1, r + n + 1, d1) - r;
    int j = lower_bound(r + 1, r + n + 1, d2) - r;
    double ans = len[j - 1] - len[i - 1];
    if (i <= n && d1 > l[i]) {
      ans -= d1 - l[i];
    }
    if (j <= n && d2 > l[j]) {
      ans += d2 - l[j];
    }
    cout << ans * (b - a) / (d2 - d1) << '\n';
  }
}
