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 = 2e5 + 5;
const double eps = 1e-9, INF = 0x3f3f3f3f;
double sy, a, b, sum[maxn], l[maxn], r[maxn];
int n, q;
inline void init() {
  scanf("%lf%lf%lf", &sy, &a, &b);
  scanf("%d", &n);
  n++;
  l[1] = r[1] = 0;
  for (int i = 2; i <= n; i++) {
    scanf("%lf%lf", &l[i], &r[i]);
    sum[i] = sum[i - 1] + (r[i] - l[i]);
  }
  ++n;
  l[n] = r[n] = INF;
  sum[n] = sum[n - 1];
}
inline void solve() {
  scanf("%d", &q);
  while (q--) {
    double x, y;
    scanf("%lf%lf", &x, &y);
    double sfb = y / (y - sy);
    double pax = a - x, pbx = b - x;
    pax *= sfb;
    pbx *= sfb;
    pax += x;
    pbx += x;
    if (pbx <= l[1] || pax >= r[n]) {
      printf("0.00000000\n");
      continue;
    }
    double ans = 0;
    int L = 1, R = n;
    while (L < R) {
      int mid = (L + R + 1) >> 1;
      if (l[mid] <= pax)
        L = mid;
      else
        R = mid - 1;
    }
    ans -= sum[L] - max(0.0, r[L] - pax);
    L = 1, R = n;
    while (L < R) {
      int mid = (L + R) >> 1;
      if (r[mid] >= pbx)
        R = mid;
      else
        L = mid + 1;
    }
    ans += sum[R];
    ans -= min(r[L] - pbx, r[L] - l[L]);
    ans = ans / sfb;
    printf("%.7f\n", ans);
  }
}
int main() {
  init();
  solve();
  return 0;
}
