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.

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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); 
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7;
long long fence[2][maxn];
long long s, a, b, pre[maxn], first, second;
int n, q;
const long double eps = 1e12;
long double getox(long long v) {
  long double val = second * (first - v);
  val /= (second - s);
  val = first - val;
  return val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> s >> a >> b >> n;
  for (int i = 1; i <= n; i++) {
    cin >> fence[0][i] >> fence[1][i];
    pre[i] = fence[1][i] - fence[0][i] + pre[i - 1];
  }
  cin >> q;
  while (q--) {
    cin >> first >> second;
    long double xl = getox(a), xr = getox(b), tot = 0;
    long long xintl = ceil(xl), xintr = floor(xr);
    int lptr = lower_bound(fence[0] + 1, fence[0] + n + 1, xintl) - fence[0];
    int rptr = upper_bound(fence[1] + 1, fence[1] + n + 1, xintr) - fence[1];
    rptr--;
    lptr--;
    if (lptr <= rptr) tot = pre[rptr] - pre[lptr];
    rptr++;
    if (lptr != rptr) {
      if (lptr >= 1 && lptr <= n && xl >= fence[0][lptr] &&
          xl <= fence[1][lptr])
        tot += (fence[1][lptr] - xl);
      if (rptr >= 1 && rptr <= n && xr >= fence[0][rptr] &&
          xr <= fence[1][rptr])
        tot += (xr - fence[0][rptr]);
    } else {
      tot = xr - xl;
    }
    tot *= (second + abs(s));
    tot /= second;
    cout.precision(12);
    cout << fixed << tot << '\n';
  }
  return 0;
}
