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 N = 400005;
double ti[N], v[N];
int A, B, n, sy, y[N], q, tot, x[N], a[N], b[N], f[N];
struct Ask {
  double x;
  int flag, from;
} ask[N * 2];
int cmp(Ask x, Ask y) { return x.x < y.x; }
int main() {
  scanf("%d%d%d%d", &sy, &A, &B, &n);
  for (int i = 1; i <= n; i++)
    scanf("%d%d", &a[i], &b[i]), f[i] = f[i - 1] + b[i] - a[i];
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &x[i], &y[i]);
    ask[++tot].flag = -1,
    ask[tot].x = A + (double)(x[i] - A) / (y[i] - sy) * (-sy);
    ask[tot].from = i;
    ask[++tot].flag = 1,
    ask[tot].x = B + (double)(x[i] - B) / (y[i] - sy) * (-sy);
    ask[tot].from = i;
    ti[i] = ask[tot].x - ask[tot - 1].x;
  }
  sort(ask + 1, ask + tot + 1, cmp);
  int now = 1;
  for (int i = 1; i <= tot; i++) {
    while (now <= n && ask[i].x >= b[now]) now++;
    double k = f[now - 1];
    if (ask[i].x >= a[now] && ask[i].x <= b[now]) k += ask[i].x - a[now];
    v[ask[i].from] += ask[i].flag * k;
  }
  for (int i = 1; i <= q; i++)
    printf("%.10lf\n", (double)v[i] / ti[i] * (B - A));
}
