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;
double f(double a, double b, double c, double d) {
  return (c - a) / (d - b) * (-b) + a;
}
int n, q;
long long s, a, b, l[200010], r[200010], t[400010], ss[400010];
int main() {
  scanf("%lld%lld%lld%d", &s, &a, &b, &n);
  for (int i = 1; i <= n; i++)
    scanf("%lld%lld", &l[i], &r[i]),
        t[i * 2 - 1] = l[i], t[i * 2] = r[i], ss[i * 2 - 1] = ss[i * 2 - 2],
                  ss[i * 2] = ss[i * 2 - 1] + r[i] - l[i];
  scanf("%d", &q);
  long long x, y;
  for (int i = 0; i < q; i++) {
    scanf("%lld%lld", &x, &y);
    double ql = f(a, s, x, y), qr = f(b, s, x, y);
    int L = 1, R = 2 * n, al = 0, ar = 2 * n + 1;
    while (L <= R) {
      int mid = (L + R) >> 1;
      if (t[mid] <= ql)
        al = mid, L = mid + 1;
      else
        R = mid - 1;
    }
    L = 1, R = 2 * n;
    while (L <= R) {
      int mid = (L + R) >> 1;
      if (t[mid] >= qr)
        ar = mid, R = mid - 1;
      else
        L = mid + 1;
    }
    double ans = 0;
    if (al + 1 == ar)
      ans += (qr - ql) * (al & 1);
    else {
      ans += ss[ar - 1] - ss[al + 1];
      if (!(ar & 1)) ans += qr - l[ar / 2];
      if (al & 1) ans += -ql + r[(al + 1) / 2];
    }
    printf("%.7lf\n", ans / y * (y - s));
  }
  return 0;
}
