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 double EPS = 1e-12;
const double DELTA = 1e-12;
const int INF = ~(1 << 31);
const long long LINF = ~(1LL << 63);
const int MOD = (int)998244353;
const int MAXN = 3e5 + 10;
long long sy, a, b, n, l, r, q, x, y;
long long a1, b1, c1, a2, b2, c2;
vector<long long> fence{};
vector<long long> sums{};
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> sy >> a >> b >> n;
  sums.push_back(0);
  fence.push_back(0);
  for (int i = 0; i < n; ++i) {
    cin >> l >> r;
    fence.push_back(l);
    fence.push_back(r);
    int len = r - l;
    sums.push_back(sums[sums.size() - 1] + len);
  }
  fence.push_back(1e9 + 10);
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> x >> y;
    a1 = y - sy;
    b1 = a - x;
    c1 = -a1 * x - b1 * y;
    double x1 = -c1 / (a1 + 0.);
    long long leftFenceIdx =
        lower_bound((fence).begin(), (fence).end(), x1) - fence.begin();
    a2 = y - sy;
    b2 = b - x;
    c2 = -a2 * x - b2 * y;
    double x2 = -c2 / (a2 + 0.);
    long long rightFenceIdx =
        prev(upper_bound((fence).begin(), (fence).end(), x2)) - fence.begin();
    double coef = x != a ? (x - a) / (x - x1) : (x - b) / (x - x2);
    double currSum = 0;
    if (leftFenceIdx % 2 == 0) {
      currSum += fence[leftFenceIdx] - x1;
    }
    if (rightFenceIdx % 2 == 1) {
      currSum += x2 - fence[rightFenceIdx];
    }
    if (leftFenceIdx % 2 == 0 && leftFenceIdx - rightFenceIdx == 1) {
      currSum -= (fence[leftFenceIdx] - fence[rightFenceIdx]);
    }
    if (rightFenceIdx - leftFenceIdx > 0) {
      long long rightSumsIdx = rightFenceIdx / 2 - 1;
      long long leftSumsIdx = leftFenceIdx / 2;
      currSum += sums[rightSumsIdx + 1] - sums[leftSumsIdx];
    }
    printf("%.10f\n", currSum * coef);
  }
  return 0;
}
