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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target( \
    "avx,avx2,fma,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
using namespace std;
const long long mod = 1000000007;
mt19937 rng((int)chrono::steady_clock::now().time_since_epoch().count());
const int mxN = 200010;
const long double EPS = 1e-9;
long double A, B, Sy;
long double L[mxN], R[mxN];
long double Psum[mxN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(15);
  int N, Q;
  long long x, y;
  cin >> Sy >> A >> B;
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> x >> y;
    L[i] = x, R[i] = y;
    Psum[i] = y - x;
    if (i) Psum[i] += Psum[i - 1];
  }
  Psum[N] = Psum[N - 1];
  cin >> Q;
  while (Q--) {
    cin >> x >> y;
    long double Ax = x + (A - x) * y / (y - Sy);
    long double Bx = x + (B - x) * y / (y - Sy);
    int itA = lower_bound(L, L + N, Ax) - L - 1;
    itA = max(itA, 0);
    while (itA < N && Ax > L[itA] + EPS) itA++;
    int itB = lower_bound(L, L + N, Bx) - L - 1;
    itB = max(itB, 0);
    while (itB < N && Bx > L[itB] + EPS) itB++;
    itB--;
    long double res = 0;
    if (itB > 0) res += Psum[itB - 1];
    if (itA > 0) res -= Psum[itA - 1];
    if (itB >= 0) res += max((long double)0, min(R[itB], Bx) - L[itB]);
    if (itA > 0) res += max((long double)0, R[itA - 1] - max(L[itA - 1], Ax));
    res *= (y - Sy) / y;
    cout << res << '\n';
  }
  return 0;
}
