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>
#pragma comment(linker, "/STACK:256000000")
#pragma warning(disable : 4996)
using namespace std;
const int inf = 1e9;
const long long llinf = 1e18;
const long double pi =
    3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825;
int main() {
  srand(time(NULL));
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(7);
  int sy, a, b;
  cin >> sy >> a >> b;
  int n;
  cin >> n;
  vector<long double> l(n), r(n);
  for (int i = 0; i < n; ++i) {
    int tmp1, tmp2;
    cin >> tmp1 >> tmp2;
    l[i] = tmp1 + 0.;
    r[i] = tmp2 + 0.;
  }
  vector<int> pref(n + 1, 0);
  for (int i = 1; i <= n; ++i) pref[i] = pref[i - 1] + r[i - 1] - l[i - 1];
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    long double l1, r1;
    l1 = a * 1. + (x - a) * 1. * (0. - sy) / (long double)(y - sy);
    r1 = b * 1. + (x - b) * 1. * (0. - sy) / (long double)(y - sy);
    int lp = lower_bound(l.begin(), l.end(), l1) - l.begin();
    int rp = upper_bound(r.begin(), r.end(), r1) - r.begin();
    long double ans = 0.;
    if (lp > 0 && r[lp - 1] > l1) ans += r[lp - 1] - l1;
    if (rp < n && r1 > l[rp]) ans += r1 - l[rp];
    if (rp - 1 >= lp) ans += (long double)(pref[rp] - pref[lp]);
    ans = min(ans, r1 - l1);
    cout << fixed << (long double)(b - a) * ans / (r1 - l1) << '\n';
  }
  return 0;
}
