#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
using namespace std;
const int N = 2e5 + 5;
pair<double, double> f[N];
double sum[N];
int main() {
  int n, q, l, r;
  double s, a, b, x, y;
  scanf("%lf %lf %lf", &s, &a, &b);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lf %lf", &f[i].first, &f[i].second);
  sum[0] = 0;
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + f[i].second - f[i].first;
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%lf %lf", &x, &y);
    double c1 = (a * y - s * x) / (y - s);
    double c2 = (b * y - s * x) / (y - s);
    int t = lower_bound(f + 1, f + n + 1, pair<double, double>(c1, 0)) - f;
    double ans = 0;
    if (t == 1)
      l = 1;
    else {
      l = t;
      if (c1 < f[t - 1].second) ans += f[t - 1].second - c1;
    }
    int tt = lower_bound(f + 1, f + n + 1, pair<double, double>(c2, 0)) - f;
    if (tt == 1)
      r = tt - 1;
    else {
      r = tt - 1;
      if (c2 < f[tt - 1].second) ans -= f[tt - 1].second - c2;
    }
    if (r >= l) ans += sum[r] - sum[l - 1];
    printf("%.10f\n", ans * (y - s) / y);
  }
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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); 

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
using namespace std;
const int N = 2e5 + 5;
pair<double, double> f[N];
double sum[N];
int main() {
  int n, q, l, r;
  double s, a, b, x, y;
  scanf("%lf %lf %lf", &s, &a, &b);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lf %lf", &f[i].first, &f[i].second);
  sum[0] = 0;
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + f[i].second - f[i].first;
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%lf %lf", &x, &y);
    double c1 = (a * y - s * x) / (y - s);
    double c2 = (b * y - s * x) / (y - s);
    int t = lower_bound(f + 1, f + n + 1, pair<double, double>(c1, 0)) - f;
    double ans = 0;
    if (t == 1)
      l = 1;
    else {
      l = t;
      if (c1 < f[t - 1].second) ans += f[t - 1].second - c1;
    }
    int tt = lower_bound(f + 1, f + n + 1, pair<double, double>(c2, 0)) - f;
    if (tt == 1)
      r = tt - 1;
    else {
      r = tt - 1;
      if (c2 < f[tt - 1].second) ans -= f[tt - 1].second - c2;
    }
    if (r >= l) ans += sum[r] - sum[l - 1];
    printf("%.10f\n", ans * (y - s) / y);
  }
  return 0;
}

```