Problem:
Score : 700 points

Problem Statement

N hotels are located on a straight line. The coordinate of the i-th hotel (1 \leq i \leq N) is x_i.

Tak the traveler has the following two personal principles:

He never travels a distance of more than L in a single day.

He never sleeps in the open. That is, he must stay at a hotel at the end of a day.

You are given Q queries. The j-th (1 \leq j \leq Q) query is described by two distinct integers a_j and b_j.
For each query, find the minimum number of days that Tak needs to travel from the a_j-th hotel to the b_j-th hotel following his principles.
It is guaranteed that he can always travel from the a_j-th hotel to the b_j-th hotel, in any given input.

Constraints

2 \leq N \leq 10^5

1 \leq L \leq 10^9

1 \leq Q \leq 10^5

1 \leq x_i < x_2 < ... < x_N \leq 10^9

x_{i+1} - x_i \leq L

1 \leq a_j,b_j \leq N

a_j \neq b_j

N,\,L,\,Q,\,x_i,\,a_j,\,b_j are integers.

Partial Score

200 points will be awarded for passing the test set satisfying N \leq 10^3 and Q \leq 10^3.

Input

The input is given from Standard Input in the following format:

N
x_1 x_2 ... x_N
L
Q
a_1 b_1
a_2 b_2
:
a_Q b_Q

Output

Print Q lines.
The j-th line (1 \leq j \leq Q) should contain the minimum number of days that Tak needs to travel from the a_j-th hotel to the b_j-th hotel.

Sample Input 1

9
1 3 6 13 15 18 19 29 31
10
4
1 8
7 3
6 7
8 5

Sample Output 1

4
2
1
2

For the 1-st query, he can travel from the 1-st hotel to the 8-th hotel in 4 days, as follows:

Day 1: Travel from the 1-st hotel to the 2-nd hotel. The distance traveled is 2.

Day 2: Travel from the 2-nd hotel to the 4-th hotel. The distance traveled is 10.

Day 3: Travel from the 4-th hotel to the 7-th hotel. The distance traveled is 6.

Day 4: Travel from the 7-th hotel to the 8-th hotel. The distance traveled is 10.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner s = new Scanner(System.in);
    int n = s.nextInt();
    int[] x = new int[n];
    for (int i = 0; i < n; i++) {
      x[i] = s.nextInt();
    }
    int limit = s.nextInt();

    int[][] dptable = new int[18][n];

    int begin = 0;
    int end = 0;
    long sum = x[0];
    while (begin < n) {
      while (true) {
        if (end+1 < n && x[end+1] - x[begin] <= limit) {
          end++;
        } else {
          break;
        }
      }
      dptable[0][begin] = end;
      begin++;
    }
    for (int i = 1; i < 18; i++) {
      for (int j = 0; j < n; j++) {
        dptable[i][j] = dptable[i-1][dptable[i-1][j]];
      }
    }
/*
      System.out.println();
    for (int i = 0; i < 18; i++) {
      for (int j = 0; j < n; j++) {
        System.out.print(dptable[i][j]);
      }
      System.out.println();
    }
      System.out.println();
*/
    int q = s.nextInt();
    for (int i = 0; i < q; i++) {
      int ta = s.nextInt() - 1;
      int tb = s.nextInt() - 1;
      int a = Math.min(ta, tb);
      int b = Math.max(ta, tb);
/*
      if (dptable[0][a] >= b) {
        System.out.println(1);
        continue;
      }
*/
      int current = a;
      int days = 0;
      int depth = 0;
      for (depth = 0; depth < 17; depth++) {
        if (dptable[depth+1][current] >= b) {
          current = dptable[depth][current];
          days = 1 << depth;
          break;
        }
      }
      while (depth >= 0) {
//        System.out.printf("%d %d\n", depth, current);
        if (dptable[depth][current] < b) {
          current = dptable[depth][current];
          days += (1 << depth);
        }
        depth--;
      }
//        System.out.printf("%d %d\n", depth, current);
      System.out.println(days + (current < b ? 1 : 0));
    }
  }
}