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.HashMap;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		
		int n = s.nextInt();
		
		int[] num = new int[n];
		int[] dp = new int[n];
		int st = 0, dr = 0;
		for (int i = 0; i < n; i++) {
			num[i] = s.nextInt();
		}
		int l = s.nextInt();
		int q = s.nextInt();
		
		
		while (dr != dp.length) {
			if (num[dr] - num[st] <= l) {
				dp[st] = dr;
				dr++;
			} else {
				if (dp[st] == 0 && st != 0)
					dp[st] = dp[st - 1];
				st++;
			}
			
			
		}
		HashMap<Integer, Integer> h = new HashMap<>();
		HashMap<Integer, Integer> h1 = new HashMap<>();
		while (st < dp.length) {
			dp[st] = dp.length - 1;
			st++;
		}
		int par = 0;
		int index = 1;
		int[] rau = new int[dp.length];
		int[] freq = new int[dp.length];
		rau[rau.length - 1] = 1;
		int apar;
		for (int i = 0; i < dp.length; i++) {
			if (rau[i] == 0) {
				par = i;
				apar = 1;
				while (par < dp.length - 1 && rau[par] == 0 || par == dp.length - 1 && freq[par] == 0) {
					rau[par] = index;
					if (freq[par] == 0)
						freq[par] = apar;
					par = dp[par];
					
					apar++;
					if (par == dp.length - 1 && freq[par] != 0) break;
				}
				h.put(index, apar);
				h1.put(index, par);
				index++;
			}
		}
		
		//System.exit(1);
		
	/*	for (int i : rau) {
			System.out.print(i);
		}
		System.out.println();
		for (int i : freq) {
			System.out.print(i);
		}*/
		for (int i = 0; i < q; i++) {
			int a = s.nextInt() - 1;
			int b = s.nextInt() - 1;
			if (a > b) {
				a = a ^ b;
				b = a ^ b;
				a = a ^ b;
			}
			//System.out.println(f(dp, a - 1, b - 1));
			int cur = rau[a];
			int res = 0;
			int dest = rau[b];
			//System.out.println(rau[b]+" "+rau[a]);
			res -= freq[a];
			int last = 0;
			
			while (cur != dest) {
				res += h.get(cur) - last;
				last = freq[h1.get(cur)];
				if(cur<rau[h1.get(cur)])System.exit(1);
				cur = rau[h1.get(cur)];
				
				//System.out.println(last);
				
			}
			//System.out.println(last);
			res += freq[b] - last;
			System.out.println(res);
		}
		
		
	}
	
	public static int f(int[] dp, int a, int b) {
		
		int cost = 0;
		if (a > b) {
			a = a ^ b;
			b = a ^ b;
			a = a ^ b;
			
		}
		int current = a;
		while (current < b) {
			current = dp[current];
			cost++;
		}
		return cost;
		
		
	}
}