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: Time Limit Exceeded)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define fre(x) freopen(x".in","r",stdin),freopen(x".out","w",stdout)
using namespace std;

inline int read(){
	int sum=0,f=1;char ch=getchar();
	while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
	return f*sum;
}

const int N=100005;
struct node{int l,r,pre,nxt;}a[N];
struct ques{int l,r,id,ans;}q[N];
int n,m,L,p[N];

inline bool cmp(ques a,ques b){
	return a.l!=b.l? a.l<b.l:a.r<b.r;
}

inline bool cmp1(ques a,ques b){return a.id<b.id;}

signed main(){
	n=read();
	for(int i=1;i<=n;i++) p[i]=read();
	L=read();m=read();
	for(int i=1;i<=m;i++) q[i].l=read(),q[i].r=read();
	for(int i=1;i<=m;i++) q[i].id=i;
	for(int i=1;i<=m;i++) if(q[i].l>q[i].r) swap(q[i].l,q[i].r);
	for(int i=1;i<=n;i++){
		int k=p[i]+L;
		k=upper_bound(p+1,p+n+1,k)-p-1;
		a[i].nxt=k;
		if(!a[k].l||i<a[k].l) a[k].l=i;
		a[k].r=max(a[k].r,i);
		a[i].pre=upper_bound(p+1,p+n+1,p[i]-L)-p;
	}
	for(int i=1;i<=m;i++){
		if(q[i].l<q[i].r){
			while(q[i].l<q[i].r) q[i].l=a[q[i].l].nxt,q[i].ans++;
		}
		else{
			while(q[i].r>q[i].l) q[i].r=a[q[i].r].pre,q[i].ans++;
		}
		
	}
	sort(q+1,q+m+1,cmp1);
	for(int i=1;i<=m;i++) printf("%d\n",q[i].ans);
	return 0;
}