#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
const int mod = 1e9 + 7;
int sy, a, b, n, q, p1, p2, x, y;
int s[N], l[N], r[N];
double L, R, ans;
int main() {
  scanf("%d%d%d%d", &sy, &a, &b, &n);
  for (int i = (1); i < (n + 1); i++)
    scanf("%d%d", &l[i], &r[i]), s[i] = s[i - 1] + r[i] - l[i];
  scanf("%d", &q);
  for (int i = (1); i < (q + 1); i++) {
    scanf("%d%d", &x, &y);
    ans = 0;
    L = x - 1.0 * y / (y - sy) * (x - a);
    R = x - 1.0 * y / (y - sy) * (x - b);
    p1 = lower_bound(l + 1, l + n + 1, L) - l;
    p2 = upper_bound(r + 1, r + n + 1, R) - r;
    p2--;
    if (p1 <= p2) ans += s[p2] - s[p1 - 1];
    if (p1 - 1 == p2 + 1 && p1 - 1 >= 1 && p1 - 1 <= n)
      ans += R - L;
    else {
      if (p1 - 1 > 0 && L >= l[p1 - 1] && L <= r[p1 - 1]) ans += r[p1 - 1] - L;
      if (p2 + 1 <= n && R >= l[p2 + 1] && R <= r[p2 + 1]) ans += R - l[p2 + 1];
    }
    ans *= 1.0 * (y - sy) / y;
    printf("%.8f\n", ans);
  }
  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 N = 200005;
const int mod = 1e9 + 7;
int sy, a, b, n, q, p1, p2, x, y;
int s[N], l[N], r[N];
double L, R, ans;
int main() {
  scanf("%d%d%d%d", &sy, &a, &b, &n);
  for (int i = (1); i < (n + 1); i++)
    scanf("%d%d", &l[i], &r[i]), s[i] = s[i - 1] + r[i] - l[i];
  scanf("%d", &q);
  for (int i = (1); i < (q + 1); i++) {
    scanf("%d%d", &x, &y);
    ans = 0;
    L = x - 1.0 * y / (y - sy) * (x - a);
    R = x - 1.0 * y / (y - sy) * (x - b);
    p1 = lower_bound(l + 1, l + n + 1, L) - l;
    p2 = upper_bound(r + 1, r + n + 1, R) - r;
    p2--;
    if (p1 <= p2) ans += s[p2] - s[p1 - 1];
    if (p1 - 1 == p2 + 1 && p1 - 1 >= 1 && p1 - 1 <= n)
      ans += R - L;
    else {
      if (p1 - 1 > 0 && L >= l[p1 - 1] && L <= r[p1 - 1]) ans += r[p1 - 1] - L;
      if (p2 + 1 <= n && R >= l[p2 + 1] && R <= r[p2 + 1]) ans += R - l[p2 + 1];
    }
    ans *= 1.0 * (y - sy) / y;
    printf("%.8f\n", ans);
  }
  return 0;
}

```