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;
double s, a, b, p[200005];
int q, n;
vector<pair<double, long long>> v;
void solve() {
  cin >> s >> a >> b;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    v.push_back({l, r});
  }
  for (int i = 0; i < n; i++) {
    p[i] = (v[i].second - v[i].first);
    if (i) p[i] += p[i - 1];
  }
  cin >> q;
  while (q--) {
    int first, second;
    double d = 0;
    cin >> first >> second;
    double l = a - s * (a - first) / (s - second),
           r = b - s * (b - first) / (s - second);
    if (l > v.back().second || r < v[0].first) {
      cout << d << '\n';
      continue;
    }
    pair<double, long long> g = {l, -1e12};
    int u = lower_bound((v).begin(), (v).end(), g) - v.begin() - 1;
    u = max(u, 0);
    g = {r, 1e12};
    int e = upper_bound((v).begin(), (v).end(), g) - v.begin() + 1;
    e = min(e, (int)v.size() - 1);
    while (u < (int)v.size() and v[u].second < l) u++;
    if (v[u].first < l) {
      d += min(r, 1.0 * v[u].second) - l;
      u++;
    }
    while (e >= 0 and v[e].first > r) e--;
    if (e >= u and v[e].second > r) {
      d += r - max(v[e].first, l);
      e--;
    }
    if (u <= e) d += p[e] - (u ? p[u - 1] : 0);
    cout << d * (second - s) / second << '\n';
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
