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 int N_ = 2E5 + 3;
double k;
int sy, X, Y;
int sum[N_], nBarrier, nPoint;
int pointfst[N_], pointsnd[N_], barrierfst[N_], barriersnd[N_];
void enter() {
  scanf("%d%d%d%d", &sy, &X, &Y, &nBarrier);
  for (int i = 1; i <= nBarrier; ++i)
    scanf("%d%d", &barrierfst[i], &barriersnd[i]);
  scanf("%d", &nPoint);
  for (int i = 1; i <= nPoint; ++i) scanf("%d%d", &pointfst[i], &pointsnd[i]);
}
double findIntersect(double x1, double afkajsdfkjkjkkk, double x2, double y2) {
  double c = (afkajsdfkjkjkkk - y2) * x1 + (x2 - x1) * afkajsdfkjkjkkk;
  double a = (afkajsdfkjkjkkk - y2);
  return c / a;
}
inline double dist(double x1, double afkajsdfkjkjkkk, double x2, double y2) {
  return sqrt((x2 - x1) * (x2 - x1) +
              (y2 - afkajsdfkjkjkkk) * (y2 - afkajsdfkjkjkkk));
}
void prep() {
  for (int i = 1; i <= nBarrier; ++i)
    sum[i] = sum[i - 1] + barriersnd[i] - barrierfst[i];
}
inline double findK(double pt, int humanfst, int humansnd) {
  double a = dist((double)humanfst, (double)humansnd, pt, 0);
  double b = dist((double)humanfst, (double)humansnd, (double)X, (double)sy);
  return a / b;
}
int binarySearchLeft(double &intersectPoint) {
  if (barriersnd[nBarrier] <= intersectPoint) return -1;
  int l = 1, r = nBarrier, mid, ret = -1;
  while (l <= r) {
    mid = (l + r) / 2;
    if (barriersnd[mid] > intersectPoint) {
      r = mid - 1;
      ret = mid;
    } else
      l = mid + 1;
  }
  return ret;
}
int binarySearchRight(double &intersectPoint) {
  if (barrierfst[1] >= intersectPoint) return -1;
  int l = 1, r = nBarrier, mid, ret = -1;
  while (l <= r) {
    mid = (l + r) / 2;
    if (barrierfst[mid] < intersectPoint) {
      l = mid + 1;
      ret = mid;
    } else
      r = mid - 1;
  }
  return ret;
}
void solve() {
  for (int i = 1; i <= nPoint; ++i) {
    double lPoint =
        findIntersect(X, sy, (double)pointfst[i], (double)pointsnd[i]);
    double rPoint =
        findIntersect(Y, sy, (double)pointfst[i], (double)pointsnd[i]);
    int lBarrier = binarySearchLeft(lPoint);
    int rBarrier = binarySearchRight(rPoint);
    if (lBarrier == -1 || rBarrier == -1 || lBarrier > rBarrier) {
      printf("0.000000\n");
      continue;
    }
    k = findK(lPoint, pointfst[i], pointsnd[i]);
    lPoint = max(lPoint, (double)barrierfst[lBarrier]);
    rPoint = min(rPoint, (double)barriersnd[rBarrier]);
    double lLeak = abs(lPoint - (double)barrierfst[lBarrier]);
    double rLeak = abs(rPoint - (double)barriersnd[rBarrier]);
    printf("%.6f\n",
           ((double)sum[rBarrier] - (double)sum[lBarrier - 1] - lLeak - rLeak) /
               k);
  }
}
int main() {
  enter();
  prep();
  solve();
  return 0;
}
