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 <cstdio>
#include <algorithm>
#include <cmath>
#define N 100002
using namespace std;
int n,i,j,a[N],dis[N],l,q,r[N],f[N][22];
int main()
{
	cin>>n;
	for(i=1;i<=n;i++) cin>>a[i];
	cin>>l>>q;
	for(i=1;i<=n;i++){
		int pos=lower_bound(a+1,a+n+1,a[i]+l)-a;
		if(a[pos]==a[i]+l) f[i][0]=pos;
		else f[i][0]=pos-1;
	}
	for(j=0;(1<<j)<=n;j++){
		for(i=1;i<=n;i++) f[i][j+1]=f[f[i][j]][j];
	}
	for(i=1;i<=q;i++){
		int x,y,ans=0;
		cin>>x>>y;
		if(x>y) swap(x,y);
		for(j=log2(1.0*n);j>=0;j--){
			if(f[x][j]<=y) ans+=pow(2,j),x=f[x][j];
			if(x==y) break;
		}
		cout<<ans<<endl;
	}
	return 0;
}