#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 200000;
pair<double, double> lrs[MAX_N];
int fss[MAX_N + 1];
int main() {
  int sy, a, b, n;
  scanf("%d%d%d%d", &sy, &a, &b, &n);
  for (int i = 0; i < n; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    lrs[i] = pair<double, double>(l, r);
    fss[i + 1] = fss[i] + (r - l);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    double xa = x + (double)(a - x) * y / (y - sy);
    double xb = x + (double)(b - x) * y / (y - sy);
    int ka = lower_bound(lrs, lrs + n, pair<double, double>(xa, xa)) - lrs;
    int kb = lower_bound(lrs, lrs + n, pair<double, double>(xb, xb)) - lrs;
    double sum = fss[kb] - fss[ka];
    if (kb > 0 && xb < lrs[kb - 1].second) sum -= (lrs[kb - 1].second - xb);
    if (ka > 0 && xa < lrs[ka - 1].second) sum += (lrs[ka - 1].second - xa);
    double t = sum * (b - a) / (xb - xa);
    printf("%.10lf\n", t);
  }
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
const int MAX_N = 200000;
pair<double, double> lrs[MAX_N];
int fss[MAX_N + 1];
int main() {
  int sy, a, b, n;
  scanf("%d%d%d%d", &sy, &a, &b, &n);
  for (int i = 0; i < n; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    lrs[i] = pair<double, double>(l, r);
    fss[i + 1] = fss[i] + (r - l);
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    double xa = x + (double)(a - x) * y / (y - sy);
    double xb = x + (double)(b - x) * y / (y - sy);
    int ka = lower_bound(lrs, lrs + n, pair<double, double>(xa, xa)) - lrs;
    int kb = lower_bound(lrs, lrs + n, pair<double, double>(xb, xb)) - lrs;
    double sum = fss[kb] - fss[ka];
    if (kb > 0 && xb < lrs[kb - 1].second) sum -= (lrs[kb - 1].second - xb);
    if (ka > 0 && xa < lrs[ka - 1].second) sum += (lrs[ka - 1].second - xa);
    double t = sum * (b - a) / (xb - xa);
    printf("%.10lf\n", t);
  }
  return 0;
}

```