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 maxn = 200000 + 10;
double lx, rx, lenx, sum[maxn];
int n, m, sy, a, b, l[maxn], r[maxn];
int find(double x) {
  if (x < l[1]) return 0;
  int L = 1, R = n + 1, mid;
  while (L + 1 < R) {
    mid = (L + R) >> 1;
    if (l[mid] <= x)
      L = mid;
    else
      R = mid;
  }
  return L;
}
int main() {
  scanf("%d%d%d%d", &sy, &a, &b, &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &l[i], &r[i]);
    if (i > 1) sum[i] = sum[i - 1] + (double)(l[i] - r[i - 1]);
  }
  scanf("%d", &m);
  for (int i = 1, x, y, L, R; i <= m; ++i) {
    scanf("%d%d", &x, &y);
    lx = (1. * x * sy - 1. * a * y) / (sy == y ? 1. : 1. * (sy - y));
    rx = (1. * x * sy - 1. * b * y) / (sy == y ? 1. : 1. * (sy - y));
    L = find(lx);
    R = find(rx);
    if (!R)
      puts("0.0000000000");
    else {
      lenx = min(rx, (double)r[R]) - max((double)l[1], min(lx, (double)r[L]));
      if (L < R) lenx -= sum[R] - sum[L];
      printf("%.10lf\n", 1. * (b - a) * lenx / (rx - lx));
    }
  }
  return 0;
}
