#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
long long sy, a, b;
long long n, q;
long long fenl[N], fenr[N];
long long sum[N];
struct point {
  long long x, y;
} p[N];
int main() {
  scanf("%lld%lld%lld", &sy, &a, &b);
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld%lld", &fenl[i], &fenr[i]);
    sum[i] = sum[i - 1] + fenr[i] - fenl[i];
  }
  scanf("%lld", &q);
  for (int i = 1; i <= q; i++) scanf("%lld%lld", &p[i].x, &p[i].y);
  for (int i = 1; i <= q; i++) {
    double ans = 0;
    double x1 = 1.0 * p[i].x - 1.0 * p[i].y * (p[i].x - a) / (p[i].y - sy);
    double x2 = 1.0 * p[i].x - 1.0 * p[i].y * (p[i].x - b) / (p[i].y - sy);
    double sp = 1.0 * p[i].y / (p[i].y - sy);
    long long L = lower_bound(fenl + 1, fenl + n + 1, x1) - fenl;
    long long R = upper_bound(fenr + 1, fenr + n + 1, x2) - fenr;
    R -= 1;
    ans += 1.0 * (sum[R] - sum[L - 1]) / sp;
    if (x1 < fenr[L - 1]) ans += 1.0 * (fenr[L - 1] - x1) / sp;
    if (R < n && x2 > fenl[R + 1]) ans += 1.0 * (x2 - fenl[R + 1]) / sp;
    printf("%.6llf\n", ans);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 N = 2e5 + 5;
long long sy, a, b;
long long n, q;
long long fenl[N], fenr[N];
long long sum[N];
struct point {
  long long x, y;
} p[N];
int main() {
  scanf("%lld%lld%lld", &sy, &a, &b);
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld%lld", &fenl[i], &fenr[i]);
    sum[i] = sum[i - 1] + fenr[i] - fenl[i];
  }
  scanf("%lld", &q);
  for (int i = 1; i <= q; i++) scanf("%lld%lld", &p[i].x, &p[i].y);
  for (int i = 1; i <= q; i++) {
    double ans = 0;
    double x1 = 1.0 * p[i].x - 1.0 * p[i].y * (p[i].x - a) / (p[i].y - sy);
    double x2 = 1.0 * p[i].x - 1.0 * p[i].y * (p[i].x - b) / (p[i].y - sy);
    double sp = 1.0 * p[i].y / (p[i].y - sy);
    long long L = lower_bound(fenl + 1, fenl + n + 1, x1) - fenl;
    long long R = upper_bound(fenr + 1, fenr + n + 1, x2) - fenr;
    R -= 1;
    ans += 1.0 * (sum[R] - sum[L - 1]) / sp;
    if (x1 < fenr[L - 1]) ans += 1.0 * (fenr[L - 1] - x1) / sp;
    if (R < n && x2 > fenl[R + 1]) ans += 1.0 * (x2 - fenl[R + 1]) / sp;
    printf("%.6llf\n", ans);
  }
  return 0;
}

```