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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<algorithm>
using namespace std;
int n,q,L,x[1<<17];
int d[20][1<<17];
main()
{
	cin>>n;
	for(int i=0;i<n;i++)cin>>x[i];
	cin>>L;
	for(int i=0;i<n;i++)
	{
		d[0][i]=upper_bound(x,x+n,x[i]+L)-x-1;
	}
	for(int k=1;k<20;k++)
	{
		for(int i=0;i<n;i++)
		{
			d[k][i]=d[k-1][d[k-1][i]];
		}
	}
	cin>>q;
	for(;q--;)
	{
		int a,b;cin>>a>>b;
		a--,b--;
		if(a>b)swap(a,b);
		int ans=0;
		for(int k=20;k--;)
		{
			if(d[k][a]<=b)
			{
				ans+=1<<k;
				a=d[k][a];
			}
		}
		cout<<ans<<endl;
	}
}