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: Time Limit Exceeded)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] x = new int[n];
        for (int i = 0; i < n; i++) {
            x[i] = scanner.nextInt();
        }

        int l = scanner.nextInt();
        int[] right = new int[n];
        int[] left = new int[n];
        for (int i = 0; i < n; i++) {
            right[i] = right(x, i, l);
            left[i] = left(x, i, l);
        }
        int q = scanner.nextInt();
        for (int i = 0; i < q; i++) {
            int a = scanner.nextInt();
            int b = scanner.nextInt();
            if (a < b) {
                System.out.println(result(right, a - 1, b - 1));
            } else {
                System.out.println(result2(left, a - 1, b - 1));
            }
        }
    }

    static int result(int[] right, int a, int b) {
        if (right[a] >= b) {
            return 1;
        }
        return 1 + result(right, right[a], b);
    }

    static int result2(int[] left, int a, int b) {
        if (left[a] <= b) {
            return 1;
        }
        return 1 + result2(left, left[a], b);
    }

    static int right(int[] x, int from, int l) {
        return binarySearch(from, x.length - 1, x, from, l);
    }

    static int left(int[] x, int from, int l) {
        return binarySearch2(0, from, x, from, l);
    }

    static int binarySearch(int begin, int end, int[] x, int from, int distance) {
        if (begin == end) {
            return begin;
        }
        if (begin + 1 == end) {
            return x[end] - x[from] <= distance ? end : begin;
        }
        int middle = (begin + end) / 2;
        if (x[middle] - x[from] < distance) {
            return binarySearch(middle, end, x, from, distance);
        } else if (x[middle] - x[from] > distance) {
            return binarySearch(begin, middle, x, from, distance);
        } else {
            return middle;
        }
    }

    static int binarySearch2(int begin, int end, int[] x, int from, int distance) {
        if (begin == end) {
            return begin;
        }
        if (begin + 1 == end) {
            return x[from] - x[begin] <= distance ? begin : end;
        }
        int middle = (begin + end) / 2;
        if (x[from] - x[middle] < distance) {
            return binarySearch2(begin, middle, x, from, distance);
        } else if (x[from] - x[middle] > distance) {
            return binarySearch2(middle, end, x, from, distance);
        } else {
            return middle;
        }
    }

}