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;
const int N = 2e5 + 10;
double L[N], R[N], qz[N];
int n, q;
int Get(double x) {
  int l = 1, r = n, ans = -1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (L[mid] <= x)
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  return ans;
}
int main() {
  double s, a, b;
  scanf("%lf%lf%lf", &s, &a, &b);
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lf%lf", &L[i], &R[i]);
  for (int i = 1; i <= n; ++i) qz[i] = qz[i - 1] + R[i] - L[i];
  scanf("%d", &q);
  for (int o = 1; o <= q; ++o) {
    double x, y;
    scanf("%lf%lf", &x, &y);
    double t1 = (a * y - s * x) / (y - s);
    double t2 = (b * y - s * x) / (y - s);
    if (t2 <= L[1] || t1 >= R[n]) {
      printf("0\n");
      continue;
    }
    int l = Get(t1), r = Get(t2);
    double ans = 0;
    ans += min(t2, R[r]) - L[r] + qz[r - 1];
    ans -= min(t1, R[l]) - L[l] + qz[l - 1];
    printf("%.12f\n", ans * (y - s) / y);
  }
  return 0;
}
