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); 
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pli = pair<ll, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vpii = vector<pii>;
template <class T>
inline bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
inline bool ckmax(T& a, const T& b) {
  return b > a ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const char nl = '\n';
const int mxN = 2e5 + 10;
const int MOD = 1e9 + 7;
const ll infLL = 1e18;
const ld eps = 1e-6;
ll sy, a, b;
int n;
pii seg[mxN];
int xi, yi;
ll p[mxN];
vector<ld> A, B;
int32_t main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> sy >> a >> b >> n;
  for (int i = (0); i < (n); i++)
    cin >> seg[i].first >> seg[i].second,
        p[i + 1] = p[i] + seg[i].second - seg[i].first,
              A.push_back(seg[i].first), B.push_back(seg[i].second);
  int q;
  cin >> q;
  auto reduce = [&](ll v) { return ld(-sy * (xi - v)) / ld(yi - sy) + v; };
  cout << fixed << setprecision(12);
  for (int i = (0); i < (q); i++) {
    cin >> xi >> yi;
    ld l = reduce(a), r = reduce(b);
    int R = upper_bound(B.begin(), B.end(), r + eps) - B.begin();
    int L = lower_bound(A.begin(), A.end(), l - eps) - A.begin();
    ld res = p[R] - p[L];
    if (L > 0 && seg[L - 1].second > l - eps) res += seg[L - 1].second - l;
    if (R < n && seg[R].first < r + eps) res += r - seg[R].first;
    res *= ld(b - a) / ld(r - l);
    cout << res << nl;
  }
}
