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.

<image>

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;
inline void proc_status() {
  ifstream t("/proc/self/status");
  cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>())
       << endl;
}
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline T read() {
  register T sum = 0, fg = 1;
  register char ch(getchar());
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') fg = -1;
  for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) - '0' + ch;
  return sum * fg;
}
const int MAXN = (int)2e5;
int n;
double Sy, A, B;
pair<double, double> seg[MAXN + 5];
inline void input() {
  scanf("%lf%lf%lf", &Sy, &A, &B);
  n = read<int>();
  for (int i = 1; i <= n; ++i) scanf("%lf%lf", &seg[i].first, &seg[i].second);
}
struct point {
  double x, y;
  point() {}
  point(double _x, double _y) : x(_x), y(_y) {}
};
inline double slope(const point &u, const point &v) {
  return (v.y - u.y) / (v.x - u.x);
}
inline double cross(const point &u, const point &v) {
  double k = slope(u, v);
  return u.x - u.y / k;
}
inline int left_bound(double pos) {
  if (pos > seg[n].second) return n + 1;
  int l = 1, r = n;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (seg[mid].second >= pos)
      r = mid;
    else
      l = mid + 1;
  }
  return r;
}
inline int right_bound(double pos) {
  if (pos < seg[1].first) return 0;
  int l = 1, r = n;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (seg[mid].first <= pos)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
inline void solve() {
  static double prefix[MAXN + 5];
  prefix[0] = 0;
  for (int i = 1; i <= n; ++i)
    prefix[i] = prefix[i - 1] + seg[i].second - seg[i].first;
  int Q = read<int>();
  while (Q--) {
    point P;
    scanf("%lf%lf", &P.x, &P.y);
    double A0 = cross(P, point(A, Sy)), B0 = cross(P, point(B, Sy));
    int l = left_bound(A0), r = right_bound(B0);
    if (l > r) {
      puts("0.0000000");
      continue;
    }
    double ans = prefix[r] - prefix[l - 1];
    ans -= max(A0 - seg[l].first, 0.0);
    ans -= max(seg[r].second - B0, 0.0);
    printf("%.7lf\n", ans / P.y * (P.y - Sy));
  }
}
int main() {
  input();
  solve();
  return 0;
}
