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: Runtime Error)

import java.util.*;

class Main {
        private static int[][] c;
        public static void main(String args[]) {
                Scanner sc = new Scanner(System.in);
                int N = sc.nextInt();
                int[] x = new int[N];
                for (int i = 0; i < N; i++) {
                        x[i] = sc.nextInt();
                }
                c = new int[20][N];
                int L = sc.nextInt(), Q = sc.nextInt();
                int idx = 0;
                while (idx < N && x[idx] - x[0] <= L) {
                        idx++;
                }
                if (idx == N) {
                        for (int i = 0; i < Q; i++) {
                                System.out.println(1);
                        }
                        return;
                }
                c[0][0] = idx - 1;
                for (int i = 1; i < N; i++) {
                        while (idx < N && x[idx] - x[i] <= L) {
                                idx++;
                        }
                        if (idx == N) {
                                while (i < N) {
                                        c[0][i] = N - 1;
                                        i++;
                                }
                        } else {
                                c[0][i] = idx - 1;
                        }
                }
                for (int i = 1; i < 17; i++) {
                        for (int j = 0; j < N; j++) {
                                c[i][j] = c[i-1][c[i-1][j]];
                        }
                }
                for (int i = 0; i < Q; i++) {
                        int A = sc.nextInt() - 1, B = sc.nextInt() - 1;
                        if (A > B) {
                                int C = A;
                                A = B;
                                B = C;
                        }
                        System.out.println(calc(A,B));
                }
        }
        public static int calc(int A, int B) {
                int idx = 0;
                while (c[idx][A] < B) {
                        idx++;
                }
                if (idx == 0) {
                        return 1;
                } else {
                        return (1 << (idx - 1)) + calc(c[idx - 1][A], B);
                }
        }
}