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;
long double s, a, b;
int n, q;
long double prefixs[200005];
pair<long double, long double> project(long double x, long double y,
                                       long double st, long double en) {
  long double star = x - ((x - st) * (((long double)(y - s)) / y));
  long double end = x - ((x - en) * (((long double)(y - s)) / y));
  return {star, end};
}
int main() {
  double r1, r2, r3;
  scanf("%lf %lf %lf %d", &r1, &r2, &r3, &n);
  s = r1;
  a = r2;
  b = r3;
  vector<pair<long double, long double> > fenc(n);
  for (int i = 0; i < n; i++) {
    long double x, y;
    scanf("%lf %lf", &r1, &r2);
    x = r1;
    y = r2;
    fenc[i] = {x, y};
    if (i == 0) {
      prefixs[0] = y - x;
    } else {
      prefixs[i] = prefixs[i - 1] + y - x;
    }
  }
  scanf("%d", &q);
  for (int Q = 0; Q < q; Q++) {
    long double x, y;
    scanf("%lf %lf", &r1, &r2);
    x = r1;
    y = r2;
    int l = 0;
    int r = n - 1;
    int lst = -1;
    while (l < r) {
      int mid = (l + r) / 2;
      if (r - l == 1) {
        mid++;
      }
      pair<long double, long double> pr =
          project(x, y, fenc[mid].first, fenc[mid].second);
      if (pr.second < a - 0.0001) {
        l = mid;
      } else if (pr.first > a + 0.0001) {
        r = mid - 1;
      } else {
        lst = mid;
        break;
      }
    }
    if (lst == -1) {
      lst = l;
    }
    while (lst != n - 1) {
      pair<long double, long double> pr =
          project(x, y, fenc[lst + 1].first, fenc[lst + 1].second);
      if (pr.second < a) {
        lst++;
      } else {
        break;
      }
    }
    l = 0;
    r = n - 1;
    int frst = -1;
    while (l < r) {
      int mid = (l + r) / 2;
      pair<long double, long double> pr =
          project(x, y, fenc[mid].first, fenc[mid].second);
      if (pr.second < b - 0.0001) {
        l = mid + 1;
      } else if (pr.first > b + 0.0001) {
        r = mid;
      } else {
        frst = mid;
        break;
      }
    }
    if (frst == -1) {
      frst = l;
    }
    while (frst != 0) {
      pair<long double, long double> pr =
          project(x, y, fenc[frst - 1].first, fenc[frst - 1].second);
      if (pr.first > b) {
        frst--;
      } else {
        break;
      }
    }
    long double ans = 0;
    int diff = frst - lst - 1;
    if (diff > 0) {
      ans += (prefixs[frst - 1] - prefixs[lst]) * (((long double)(y - s)) / y);
    }
    pair<long double, long double> pr =
        project(x, y, fenc[lst].first, fenc[lst].second);
    if (pr.first < a + 0.0001 && pr.second > a - 0.00001) {
      if (pr.second > b - 0.0001) {
        ans += b - a;
      } else {
        ans += pr.second - a;
      }
    } else if (pr.first > a - 0.0001 && pr.first < b + 0.0001) {
      if (pr.second > b - 0.0001) {
        ans += b - pr.first;
      } else {
        ans += pr.second - pr.first;
      }
    }
    if (frst != lst) {
      pr = project(x, y, fenc[frst].first, fenc[frst].second);
      if (pr.first < b + 0.0001 && pr.second > b - 0.0001) {
        if (pr.first < a + 0.0001) {
          ans += b - a;
        } else {
          ans += b - pr.first;
        }
      } else if (pr.second < b + 0.0001 && pr.second > a - 0.0001) {
        if (pr.first < a + 0.0001) {
          ans += pr.second - a;
        } else {
          ans += pr.second - pr.first;
        }
      }
    }
    printf("%.15Lf\n", ans);
  }
  return 0;
}
