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;
int n, m, q, l, r;
double d, s, t, x, y;
int tree[2000010 * 4], pl[2000010 * 4], pr[2000010 * 4], cnt = 1;
bool flag[2000010 * 4];
void change(int l, int r, int id, int l1, int r1) {
  if (l >= l1 && r <= r1) {
    tree[id] = r - l + 1;
    flag[id] = 1;
    return;
  }
  if (flag[id]) return;
  int mid = (l + r) >> 1;
  if (l1 <= mid) {
    if (pl[id] == 0) pl[id] = ++cnt;
    change(l, mid, pl[id], l1, r1);
  }
  if (r1 > mid) {
    if (pr[id] == 0) pr[id] = ++cnt;
    change(mid + 1, r, pr[id], l1, r1);
  }
  tree[id] = tree[pl[id]] + tree[pr[id]];
}
int dist(int l, int r, int l1, int r1) {
  int l2 = max(l, l1);
  int r2 = min(r, r1);
  return r2 - l2 + 1;
}
int query(int l, int r, int id, int l1, int r1) {
  if (l1 > r1) return 0;
  if (l >= l1 && r <= r1) return tree[id];
  if (flag[id]) return dist(l, r, l1, r1);
  int mid = (l + r) >> 1;
  int res = 0;
  if (l1 <= mid && pl[id]) res += query(l, mid, pl[id], l1, r1);
  if (r1 > mid && pr[id]) res += query(mid + 1, r, pr[id], l1, r1);
  return res;
}
int main() {
  scanf("%lf%lf%lf", &d, &s, &t);
  d *= -1.0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &l, &r);
    change(1, 0x3FFFFFFF, 1, l, r - 1);
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    scanf("%lf%lf", &x, &y);
    double ts = (y + d) / y;
    double l1 = x + (s - x) / ts;
    double r1 = x + (t - x) / ts;
    int l2 = ceil(l1);
    int r2 = floor(r1);
    double ans = 0;
    if (l2 == r2 + 1) {
      if (query(1, 0x3FFFFFFF, 1, l2 - 1, l2 - 1)) ans = r1 - l1;
    } else {
      ans += query(1, 0x3FFFFFFF, 1, l2, r2 - 1);
      if (query(1, 0x3FFFFFFF, 1, l2 - 1, l2 - 1)) ans += l2 * 1.0 - l1;
      if (query(1, 0x3FFFFFFF, 1, r2, r2)) ans += r1 - r2 * 1.0;
    }
    ans *= ts;
    printf("%.6lf\n", ans);
  }
}
