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;
struct chash {
  long long operator()(long long x) const { return hash<long long>{}(x); }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long sy, a, b;
  cin >> sy >> a >> b;
  long long n;
  cin >> n;
  long long X[n], Y[n];
  long long sum[n];
  for (long long i = 0; i < n; i++) cin >> X[i] >> Y[i], sum[i] = Y[i] - X[i];
  for (long long i = 1; i < n; i += 1) sum[i] += sum[i - 1];
  long long q;
  cin >> q;
  cout << setprecision(6) << fixed;
  while (q--) {
    long long x, y;
    cin >> x >> y;
    long double x1 = x + (1.0 * y * (x - a) / (sy - y));
    long double x2 = x + (1.0 * y * (x - b) / (sy - y));
    long long l1 = lower_bound(Y, Y + n, x1) - Y;
    long long l2 = lower_bound(Y, Y + n, x2) - Y;
    long double left = 0, right = 0;
    left =
        ((l1 - 1) >= 0 ? sum[l1 - 1] : 0) +
        ((l1 < n) ? min((long double)Y[l1], max(x1, (long double)X[l1])) - X[l1]
                  : 0);
    right =
        ((l2 - 1) >= 0 ? sum[l2 - 1] : 0) +
        ((l2 < n) ? min((long double)Y[l2], max(x2, (long double)X[l2])) - X[l2]
                  : 0);
    cout << ((right - left) * ((long double)(y - sy) / (y))) << "\n";
  }
  return 0;
}
