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;
int n, q, plen;
double ans, sy, a, b, x, y, a_, b_;
pair<int, pair<int, int>> f[400005];
bool cm0(int i, decltype(*f)& p) { return i < p.first; }
bool cm1(decltype(*f)& p, int i) { return p.first < i; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> sy >> a >> b;
  cin >> n;
  n <<= 1;
  for (int i = 0; i < n; ++i) cin >> f[i].first, f[i].second.second = !(i & 1);
  sort(f, f + n);
  f[1].second.first = f[1].first - f->first;
  for (int i = 1; i < n / 2; ++i) {
    f[i << 1].second.first = f[i * 2 - 1].second.first;
    f[i << 1 | 1].second.first =
        f[i << 1].second.first + f[i << 1 | 1].first - f[i << 1].first;
  }
  for (cin >> q; q--;) {
    cin >> x >> y;
    ans = 0;
    a_ = x + (a - x) * y / (y - sy);
    b_ = x + (b - x) * y / (y - sy);
    if (a_ >= f[n - 1].first || b_ <= f->first) {
      puts("0");
      continue;
    }
    auto it = upper_bound(f, f + n, a_, cm0);
    if (!it->second.second) ans += it->first - a_;
    auto jt = lower_bound(it, f + n, ceil(b_), cm1);
    --jt;
    if (jt->second.second) ans += b_ - jt->first;
    ans += jt->second.first - it->second.first;
    printf("%.8lf\n", ans * (b - a) / (b_ - a_));
  }
}
