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;
void preproc() {}
void reset() {}
void solve() {
  long long sy, a, b;
  cin >> sy >> a >> b;
  int n;
  cin >> n;
  vector<pair<long long, long long>> points(n);
  for (auto &p : points) cin >> p.first >> p.second;
  vector<long long> pref(2 * n + 2), key(2 * n + 2);
  key[0] = 0;
  key[2 * n + 1] = (int)1e9 + 2;
  for (int x = 0; x < n; ++x) {
    key[2 * x + 1] = points[x].first, key[2 * x + 2] = points[x].second;
    pref[2 * x + 2] = points[x].second - points[x].first;
  }
  for (int x = 1; x <= 2 * n + 1; ++x) pref[x] += pref[x - 1];
  ;
  ;
  auto OX = [&](long double xi, long double yi, long double x,
                long double y) -> long double {
    return -y * (xi - x) / (yi - y) + x;
  };
  int q;
  cin >> q;
  while (q--) {
    int xi, yi;
    cin >> xi >> yi;
    long double Ix = OX(xi, yi, a, sy);
    long double Iy = OX(xi, yi, b, sy);
    long long Ixr = ceil(Ix);
    long long Iyr = floor(Iy);
    auto itx = lower_bound(key.begin(), key.end(), Ixr);
    auto ity = upper_bound(key.begin(), key.end(), Iyr);
    ity = prev(ity);
    int j_x = (int)(itx - key.begin());
    int j_y = (int)(ity - key.begin());
    ;
    ;
    long double res = 0;
    if (j_x <= j_y) {
      res = pref[j_y] - pref[j_x] + ((j_x - 1) & 1 ? key[j_x] - Ix : 0) +
            ((j_y - 1) % 2 == 0 ? Iy - key[j_y] : 0);
    } else
      res = ((j_x - 1) & 1 ? Iy - Ix : 0);
    cout << res * (b - a) / (Iy - Ix) << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  preproc();
  int tc = 1;
  for (int Tt = 1; Tt <= tc; ++Tt) {
    reset();
    solve();
  }
  return EXIT_SUCCESS;
}
