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.

<image>

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;
int n, q, lo, hi, mid, A, B, l[1 << 20], p[1 << 20], sum[1 << 20], rx, ry, rs,
    ra, rb;
double k, lewy, prawy, s, a, b, res, x, y;
int lbound(int x) {
  lo = 1;
  hi = n + 1;
  mid = (lo + hi) / 2;
  while (lo != hi) {
    if (l[mid] > x)
      hi = mid;
    else
      lo = mid + 1;
    mid = (lo + hi) / 2;
  }
  return mid - 1;
}
double przec(double pocz, double kon) {
  if (pocz > prawy || kon < lewy) return 0;
  pocz = max(pocz, lewy);
  kon = min(kon, prawy);
  return kon - pocz;
}
int main() {
  scanf("%d%d%d", &rs, &ra, &rb);
  s = rs;
  a = ra;
  b = rb;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l[i], &p[i]);
    sum[i] = sum[i - 1] + (p[i] - l[i]);
  }
  scanf("%d", &q);
  l[n + 1] = INT_MAX;
  while (q--) {
    scanf("%d%d", &rx, &ry);
    x = rx;
    y = ry;
    res = 0;
    k = (y + abs(s)) / y;
    lewy = x + (a - x) / k;
    prawy = x + (b - x) / k;
    A = lbound(lewy);
    B = lbound(prawy);
    if (A > 0) res += przec(l[A], p[A]) * k;
    if (A != B && B > 0) res += przec(l[B], p[B]) * k;
    A++;
    B--;
    if (A <= B) res += (double)(sum[B] - sum[A - 1]) * k;
    printf("%f\n", res);
  }
}
