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;
using LL = long long;
const LL INF = 0x3f3f3f3f;
const LL INF64 = 0x3f3f3f3f3f3f3f3f;
const int N = 5e4 + 10;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  int sy, a, b;
  cin >> sy >> a >> b;
  int n;
  cin >> n;
  vector<pair<int, int>> segs(n);
  vector<LL> pre1(n), pre2(n);
  for (int i = 0; i < n; ++i) {
    cin >> segs[i].first >> segs[i].second;
    int tmp = segs[i].first - segs[i].second;
    pre1[i] = (i == 0 ? 0 : pre1[i - 1]) + tmp;
    pre2[i] = (i == 0 ? 0 : pre2[i - 1]) - tmp;
  }
  auto get_range = [](const vector<LL> &pre, int l, int r) {
    if (l > r) return 0LL;
    return pre[r] - (l == 0 ? 0 : pre[l - 1]);
  };
  auto fit = [&](double x) {
    if (x > b)
      x = b;
    else if (x < a)
      x = a;
    return x;
  };
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    auto f = [&](LL x2) { return 1.0 * (sy * (x - x2) + y * x2) / y; };
    int l = 0, r = n - 1;
    while (l < r) {
      int mid = r - (r - l) / 2;
      int x2 = segs[mid].first;
      if (f(x2) < a)
        l = mid;
      else
        r = mid - 1;
    }
    int t1 = l + 1;
    double ans = 0;
    ans += fit(f(segs[l].second)) - fit(f(segs[l].first));
    l = l + 1;
    r = n - 1;
    while (l < r) {
      int mid = l + (r - l) / 2;
      int x2 = segs[mid].second;
      if (f(x2) > b)
        r = mid;
      else
        l = mid + 1;
    }
    int t2 = l - 1;
    if (l < n) ans += fit(f(segs[l].second)) - fit(f(segs[l].first));
    ans += 1.0 * sy * get_range(pre1, t1, t2) / y + get_range(pre2, t1, t2);
    cout << ans << '\n';
  }
  return 0;
}
