#include <bits/stdc++.h>
using namespace std;
vector<pair<double, double> > v;
vector<double> tot;
double cal(double first) {
  auto it = lower_bound(v.begin(), v.end(), make_pair((first), (0.0)));
  double ret = 0;
  if (it != v.begin()) {
    it--;
    ret += tot[it - v.begin()];
    ret += min(it->second, first) - it->first;
  }
  return ret;
}
int main() {
  int s, a, b;
  scanf("%d %d %d", &s, &a, &b);
  int n;
  scanf("%d", &n);
  double last = 0;
  tot.push_back(0);
  for (int i = 0; i < n; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    v.push_back(make_pair((l), (r)));
    last += r - l;
    tot.push_back(last);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int first, second;
    scanf("%d %d", &first, &second);
    double l = a + (double)(first - a) / (second - s) * (-s),
           r = b + (double)(first - b) / (second - s) * (-s);
    double tot = cal(r) - cal(l);
    printf("%.12lf\n", tot / (r - l) * (b - a));
  }
}

### Prompt

Create a solution in cpp for 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>
using namespace std;
vector<pair<double, double> > v;
vector<double> tot;
double cal(double first) {
  auto it = lower_bound(v.begin(), v.end(), make_pair((first), (0.0)));
  double ret = 0;
  if (it != v.begin()) {
    it--;
    ret += tot[it - v.begin()];
    ret += min(it->second, first) - it->first;
  }
  return ret;
}
int main() {
  int s, a, b;
  scanf("%d %d %d", &s, &a, &b);
  int n;
  scanf("%d", &n);
  double last = 0;
  tot.push_back(0);
  for (int i = 0; i < n; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    v.push_back(make_pair((l), (r)));
    last += r - l;
    tot.push_back(last);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int first, second;
    scanf("%d %d", &first, &second);
    double l = a + (double)(first - a) / (second - s) * (-s),
           r = b + (double)(first - b) / (second - s) * (-s);
    double tot = cal(r) - cal(l);
    printf("%.12lf\n", tot / (r - l) * (b - a));
  }
}

```