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 = 1000005;
struct frog {
  int l, len;
};
frog a[N];
double diy, hap, you;
int m, hey, sum[N];
int gt(int l, int r) {
  if (l <= r)
    return sum[r] - sum[l - 1];
  else
    return 0;
}
double gr(double x, double y) {
  if (x == you) return you;
  double k = (y - diy) / (x - you);
  return x - y / k;
}
double gl(double x, double y) {
  if (x == hap) return hap;
  double k = (y - diy) / (x - hap);
  return x - y / k;
}
int lower(int x) {
  int l = 1, r = m, ans = 0;
  while (l <= r) {
    int mid = l + r >> 1;
    if (a[mid].l <= x) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  return ans;
}
void pd(double x, double y) {
  int bbb, ccc;
  double l = gl(x, y), r = gr(x, y), aaa1, aaa;
  bbb = lower((int)l);
  ccc = lower((int)r);
  if (bbb == ccc) {
    if (a[bbb].l <= l && a[ccc].l + a[ccc].len >= r) {
      aaa1 = r - l;
      printf("%.15lf\n", aaa1 / y * (y - diy));
      return;
    }
  }
  aaa = gt(bbb + 1, ccc - 1);
  if (1.0 * a[bbb].l + 1.0 * a[bbb].len > l)
    aaa += (a[bbb].l * 1.0 + a[bbb].len * 1.0 - l);
  if (1.0 * a[ccc].l + 1.0 * a[ccc].len > r)
    aaa += r - 1.0 * a[ccc].l;
  else if (a[ccc].l > l)
    aaa += 1.0 * a[ccc].len;
  printf("%.15lf\n", aaa / y * (y - diy));
  return;
}
int x, y, iq;
int main() {
  scanf("%lf%lf%lf%d", &diy, &hap, &you, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i].l, &a[i].len);
    a[i].len -= a[i].l;
  }
  for (int i = 1; i <= m; i++) sum[i] = sum[i - 1] + a[i].len;
  scanf("%d", &iq);
  for (int i = 1; i <= iq; i++) {
    scanf("%d%d", &x, &y);
    pd(x, y);
  }
  return 0;
}
