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;
const int maxN = 200011;
const double eps = 1e-6;
struct segm {
  double a, b;
};
struct point {
  double x, y;
};
struct line {
  double a, b, c;
  double eval(point P) { return a * P.x + b * P.y + c; }
  void from_points(point A, point B) {
    a = B.y - A.y;
    b = A.x - B.x;
    c = -(a * A.x + b * A.y);
  }
  point operator+(line B) {
    line A = *this;
    double upx = A.b * B.c - A.c * B.b;
    double upy = A.a * B.c - A.c * B.a;
    double down = A.a * B.b - A.b * B.a;
    double x = upx / down;
    double y = -upy / down;
    return {x, y};
  }
};
double sy, a, b;
int n, i, q;
segm S[maxN];
point P[maxN];
line base;
double len[maxN];
int find_first(point P) {
  int ans = 0;
  double act;
  line aux;
  for (int step = 1 << 18; step > 0; step >>= 1) {
    if (ans + step > n) continue;
    act = S[ans + step].b;
    aux.from_points(P, {act, 0});
    act = (aux + base).x;
    if (act < a + eps) ans += step;
  }
  return ans + 1;
}
int find_last(point P) {
  int ans = 0;
  double act;
  line aux;
  for (int step = 1 << 18; step > 0; step >>= 1) {
    if (ans + step > n) continue;
    act = S[ans + step].a;
    aux.from_points(P, {act, 0});
    act = (aux + base).x;
    if (act <= b + eps) ans += step;
  }
  return ans;
}
double get_inters(point P, segm S) {
  double act, p1, p2;
  line aux;
  act = S.a;
  aux.from_points(P, {act, 0});
  act = (aux + base).x;
  p1 = max(a, act);
  act = S.b;
  aux.from_points(P, {act, 0});
  act = (aux + base).x;
  p2 = min(b, act);
  return max(0.00, p2 - p1);
}
double solve(point P) {
  int bg = find_first(P);
  int en = find_last(P);
  if (bg > en) return 0;
  double sol = get_inters(P, S[bg]);
  if (bg != en) sol += get_inters(P, S[en]);
  double coef = (P.y - sy) / P.y;
  if (en - bg >= 2) sol += (len[en - 1] - len[bg]) * coef;
  return sol;
}
int main() {
  scanf("%lf%lf%lf", &sy, &a, &b);
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%lf%lf", &S[i].a, &S[i].b);
  scanf("%d", &q);
  for (i = 1; i <= q; i++) scanf("%lf%lf", &P[i].x, &P[i].y);
  base.from_points({a, sy}, {b, sy});
  for (i = 1; i <= n; i++) {
    len[i] = S[i].b - S[i].a;
    len[i] += len[i - 1];
  }
  for (i = 1; i <= q; i++) {
    double ans = solve(P[i]);
    printf("%.20f\n", ans);
  }
  return 0;
}
