#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 10;
const int MOD = 1e9 + 7;
const double PI = acos(-1.0);
long long L[MAX], R[MAX];
long long sum[MAX];
int n;
double slove(double l, double r) {
  double tot = 0;
  int x = lower_bound(R + 1, R + n + 1, l) - R;
  if (x > n) return 0;
  int y = lower_bound(R + 1, R + n + 1, r) - R;
  if (y > n) {
    y = n;
    r = R[n];
  }
  if (x == y) {
    if (r < L[y]) return 0;
    if (l >= L[y]) return r - l;
    return r - L[y];
  }
  tot += sum[y - 1] - sum[x];
  tot += max(0.0, r - L[y]);
  tot += R[x] - max(L[x] * 1.0, l);
  return tot;
}
int main() {
  long long sy, a, b;
  cin >> sy >> a >> b;
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &L[i], &R[i]);
  for (int i = 1; i <= n; i++) sum[i] += R[i] - L[i] + sum[i - 1];
  int q;
  cin >> q;
  while (q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    double l = x - 1.0 * y * (x - a) / (y - sy);
    double r = x - 1.0 * y * (x - b) / (y - sy);
    double ans = slove(l, r);
    printf("%.15lf\n", ans * (y - sy) / y);
  }
  return 0;
}

### Prompt

Your challenge is to write 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 = 2e5 + 10;
const int MOD = 1e9 + 7;
const double PI = acos(-1.0);
long long L[MAX], R[MAX];
long long sum[MAX];
int n;
double slove(double l, double r) {
  double tot = 0;
  int x = lower_bound(R + 1, R + n + 1, l) - R;
  if (x > n) return 0;
  int y = lower_bound(R + 1, R + n + 1, r) - R;
  if (y > n) {
    y = n;
    r = R[n];
  }
  if (x == y) {
    if (r < L[y]) return 0;
    if (l >= L[y]) return r - l;
    return r - L[y];
  }
  tot += sum[y - 1] - sum[x];
  tot += max(0.0, r - L[y]);
  tot += R[x] - max(L[x] * 1.0, l);
  return tot;
}
int main() {
  long long sy, a, b;
  cin >> sy >> a >> b;
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &L[i], &R[i]);
  for (int i = 1; i <= n; i++) sum[i] += R[i] - L[i] + sum[i - 1];
  int q;
  cin >> q;
  while (q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    double l = x - 1.0 * y * (x - a) / (y - sy);
    double r = x - 1.0 * y * (x - b) / (y - sy);
    double ans = slove(l, r);
    printf("%.15lf\n", ans * (y - sy) / y);
  }
  return 0;
}

```