#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200051;
const double INF = 2e9;
struct Node {
  double l, r;
  Node(double l = 0, double r = 0) : l(l), r(r) {}
} l[MAXN];
int n, q;
int sy, a, b;
long long s[MAXN];
bool cmpl(Node u, Node v) { return u.l < v.l; }
bool cmpr(Node u, Node v) { return u.r < v.r; }
int main() {
  scanf("%d%d%d", &sy, &a, &b);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    l[i] = Node(u, v);
    s[i] = s[i - 1] + (v - u);
  }
  l[n + 1] = Node(INF, INF);
  scanf("%d", &q);
  while (q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    double lx = (1.0 * v * a - 1.0 * sy * u) / (v - sy);
    double rx = (1.0 * v * b - 1.0 * sy * u) / (v - sy);
    int lp = lower_bound(l + 1, l + n + 1, Node(lx, lx), cmpl) - l - 1;
    int rp = upper_bound(l + 1, l + n + 1, Node(rx, rx), cmpr) - l;
    double ans = (lp == rp) ? (rx - lx)
                            : (max(0.0, l[lp].r - lx) + max(0.0, rx - l[rp].l) +
                               s[rp - 1] - s[lp]);
    ans = ans * (v - sy) / v;
    printf("%.10f\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 MAXN = 200051;
const double INF = 2e9;
struct Node {
  double l, r;
  Node(double l = 0, double r = 0) : l(l), r(r) {}
} l[MAXN];
int n, q;
int sy, a, b;
long long s[MAXN];
bool cmpl(Node u, Node v) { return u.l < v.l; }
bool cmpr(Node u, Node v) { return u.r < v.r; }
int main() {
  scanf("%d%d%d", &sy, &a, &b);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    l[i] = Node(u, v);
    s[i] = s[i - 1] + (v - u);
  }
  l[n + 1] = Node(INF, INF);
  scanf("%d", &q);
  while (q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    double lx = (1.0 * v * a - 1.0 * sy * u) / (v - sy);
    double rx = (1.0 * v * b - 1.0 * sy * u) / (v - sy);
    int lp = lower_bound(l + 1, l + n + 1, Node(lx, lx), cmpl) - l - 1;
    int rp = upper_bound(l + 1, l + n + 1, Node(rx, rx), cmpr) - l;
    double ans = (lp == rp) ? (rx - lx)
                            : (max(0.0, l[lp].r - lx) + max(0.0, rx - l[rp].l) +
                               s[rp - 1] - s[lp]);
    ans = ans * (v - sy) / v;
    printf("%.10f\n", ans);
  }
  return 0;
}

```