#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0;
  bool t = false;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') t = true, ch = getchar();
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - 48, ch = getchar();
  return t ? -x : x;
}
double Sy, A, B, L[200200], R[200200], p[200200];
int n, q;
int main() {
  Sy = read();
  A = read();
  B = read();
  n = read();
  for (int i = 1; i <= n; ++i) L[i] = read(), R[i] = read();
  for (int i = 1; i <= n; ++i) p[i] = p[i - 1] + R[i] - L[i];
  q = read();
  while (q--) {
    double x = read(), y = read();
    double xl = A + (x - A) * (-Sy) / (y - Sy);
    double xr = B + (x - B) * (-Sy) / (y - Sy);
    int x1 = upper_bound(&L[1], &L[n + 1], xl) - L - 1;
    int x2 = upper_bound(&L[1], &L[n + 1], xr) - L - 1;
    double lf = 0, rt = 0;
    if (x1) lf = p[x1 - 1] + min(R[x1], xl) - L[x1];
    if (x2) rt = p[x2 - 1] + min(R[x2], xr) - L[x2];
    printf("%.10lf\n", (rt - lf) * (y - Sy) / y);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
inline int read() {
  int x = 0;
  bool t = false;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') t = true, ch = getchar();
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - 48, ch = getchar();
  return t ? -x : x;
}
double Sy, A, B, L[200200], R[200200], p[200200];
int n, q;
int main() {
  Sy = read();
  A = read();
  B = read();
  n = read();
  for (int i = 1; i <= n; ++i) L[i] = read(), R[i] = read();
  for (int i = 1; i <= n; ++i) p[i] = p[i - 1] + R[i] - L[i];
  q = read();
  while (q--) {
    double x = read(), y = read();
    double xl = A + (x - A) * (-Sy) / (y - Sy);
    double xr = B + (x - B) * (-Sy) / (y - Sy);
    int x1 = upper_bound(&L[1], &L[n + 1], xl) - L - 1;
    int x2 = upper_bound(&L[1], &L[n + 1], xr) - L - 1;
    double lf = 0, rt = 0;
    if (x1) lf = p[x1 - 1] + min(R[x1], xl) - L[x1];
    if (x2) rt = p[x2 - 1] + min(R[x2], xr) - L[x2];
    printf("%.10lf\n", (rt - lf) * (y - Sy) / y);
  }
  return 0;
}

```