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:102400000,102400000")
using namespace std;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (int i = 0; i < v.size(); i++) os << v[i] << " ";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  for (typename set<T>::iterator it = v.begin(); it != v.end(); it++)
    os << *it << " ";
}
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
int gi() {
  int w = 0;
  bool q = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') q = 0, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return q ? w : -w;
}
int l[N], r[N], s[N];
int sy, a, b;
int x, y;
int main() {
  sy = gi();
  a = gi();
  b = gi();
  int n;
  n = gi();
  for (int i = (1); i <= (n); i++) {
    l[i] = gi(), r[i] = gi();
    s[i] = s[i - 1] + r[i] - l[i];
  }
  int q;
  q = gi();
  while (q--) {
    x = gi(), y = gi();
    long double rat = (long double)sy / (sy - y);
    long double ax = (x - a) * rat + a;
    long double bx = (x - b) * rat + b;
    int pa = upper_bound(l + 1, l + n + 1, ax) - l;
    int push_back = upper_bound(l + 1, l + n + 1, bx) - l;
    long double suma = s[pa - 1] - (ax < r[pa - 1] ? r[pa - 1] - ax : 0);
    long double sumb =
        s[push_back - 1] - (bx < r[push_back - 1] ? r[push_back - 1] - bx : 0);
    printf("%.9lf\n", (double)(sumb - suma) / (double)(1 - rat));
  }
  return 0;
}
