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;
double overlap(pair<int, int> iv, double lo, double hi) {
  return min(1.0 * iv.second, hi) - max(1.0 * iv.first, lo);
}
bool intersect(pair<int, int> iv, double lo, double hi) {
  return overlap(iv, lo, hi) >= 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(15);
  int sy, a, b, n;
  cin >> sy >> a >> b >> n;
  sy *= -1;
  vector<pair<int, int> > seg(n);
  for (int i = 0; i < (int)n; i++) cin >> seg[i].first >> seg[i].second;
  vector<long long> slp(n + 1);
  for (int i = 0; i < (int)n; i++)
    slp[i + 1] = slp[i] + seg[i].second - seg[i].first;
  int q;
  cin >> q;
  while (q--) {
    int first, second;
    cin >> first >> second;
    double lo = first - 1.0 * (first - a) * second / (second + sy);
    double hi = first - 1.0 * (first - b) * second / (second + sy);
    auto itlo = lower_bound(seg.begin(), seg.end(), pair<int, int>{lo, -2e9});
    if (itlo != seg.begin() && intersect(*(itlo - 1), lo, hi)) itlo--;
    if (!intersect(*itlo, lo, hi)) {
      cout << "0\n";
      continue;
    }
    auto ithi = upper_bound(seg.begin(), seg.end(), pair<int, int>{hi, 2e9});
    assert(ithi != seg.begin());
    ithi--;
    assert(intersect(*ithi, lo, hi));
    double ans = 0;
    if (ithi > itlo)
      ans += slp[ithi - seg.begin()] - slp[itlo - seg.begin() + 1];
    ans += overlap(*itlo, lo, hi);
    if (itlo != ithi) ans += overlap(*ithi, lo, hi);
    cout << ans * (second + sy) / second << "\n";
  }
}
