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 read();
int n, m, LS, RS, S;
struct POINT {
  int x, y;
} p[400021], ST, ED;
int s[400021];
int sum[400021];
long double loc(POINT a, POINT b) {
  return a.x - a.y * ((long double)a.x - b.x) / (a.y - b.y);
}
int find(long double v) {
  if (v < s[1]) return 0;
  int L = 1, R = 2 * n;
  while (L != R) {
    int mid = (L + R + 1) / 2;
    if (v - s[mid] > -1e-9) {
      L = mid;
    } else {
      R = mid - 1;
    }
  }
  return L;
}
long double len(long double L, long double R) {
  int l = find(L), r = find(R);
  long double ret = 0.000000;
  if (l & 1) {
    ret += s[l + 1] - L;
  }
  if (r & 1) {
    ret += R - s[r];
  }
  return ret + sum[r] - sum[l + 1];
}
int main() {
  scanf("%d", &S);
  LS = read();
  RS = read();
  ST.x = LS;
  ST.y = S;
  ED.x = RS;
  ED.y = S;
  n = read();
  for (int i = 1; i <= n; i++) {
    s[i * 2 - 1] = read();
    s[i * 2] = read();
    sum[i * 2 - 1] = sum[i * 2 - 2];
    sum[i * 2] = sum[i * 2 - 1] + s[i * 2] - s[i * 2 - 1];
  }
  sum[2 * n + 1] = sum[2 * n];
  m = read();
  for (int i = 1; i <= m; i++) {
    p[i].x = read();
    p[i].y = read();
    long double L = loc(ST, p[i]), R = loc(ED, p[i]);
    printf("%.15lf\n", (double)((RS - LS) / (R - L) * len(L, R)));
  }
  return 0;
}
int read() {
  int ret = 0;
  char ch = getchar();
  while (!(ch <= '9' && ch >= '0')) ch = getchar();
  while (ch <= '9' && ch >= '0') {
    ret *= 10;
    ret += ch - '0';
    ch = getchar();
  }
  return ret;
}
