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 <cstring>
#include <algorithm>
#include <cmath>
#define int long long
#define fre(x) freopen(x".in","r",stdin),freopen(x".out","w",stdout)
using namespace std;
const int BufferSize=1<<16;
char buffer[BufferSize],*head,*tail;

inline char Getchar() {
    if(head==tail) {
        int l=fread(buffer,1,BufferSize,stdin);
        tail=(head=buffer)+l;
    }
    return *head++;
}
inline int read() {
    int x=0,f=1;char c=Getchar();
    for(;!isdigit(c);c=Getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=Getchar()) x=x*10+c-'0';
    return x*f;
}

inline int exread(){
	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=200005;
struct node{int pre,nxt,to2,sp2,to1,sp1;}a[N];
struct ques{int l,r,ans;}q[N];
int n,m,L,p[N];
int bl[N],gap;

inline int work2(int l,int r){
	int ans=0;
	while(bl[l]!=bl[r]&&bl[l]!=bl[n]) ans+=a[l].sp2,l=a[l].to2;
	while(l<r) ans++,l=a[l].nxt;
	return ans;
}

inline int work1(int l,int r){
	int ans=0;
	while(bl[l]!=bl[r]&&bl[l]!=bl[1]) ans+=a[l].sp1,l=a[l].to1;
	while(l>r) ans++,l=a[l].pre;
	return ans;
}

signed main(){
//	freopen("test.in","r",stdin);
//	freopen("B.out","w",stdout);
	n=read();gap=500;
	for(int i=1;i<=n;i++) p[i]=read();
	for(int i=1;i<=n;i++) bl[i]=(i-1)/gap+1;
	L=read();m=read();
	for(int i=1;i<=m;i++) q[i].l=read(),q[i].r=read();
	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;
		a[i].pre=lower_bound(p+1,p+n+1,p[i]-L)-p;
	}
	for(int i=1;i<=n;i++){
		int p=i;
		while(bl[p]==bl[i]&&bl[p]!=bl[n]) p=a[p].nxt,a[i].sp2++;
		a[i].to2=p;
		p=i;
		while(bl[p]==bl[i]&&bl[p]!=bl[1]) p=a[p].pre,a[i].sp1++;
		a[i].to1=p;
	}
	for(int i=1;i<=m;i++){
		if(q[i].l<q[i].r) q[i].ans=work2(q[i].l,q[i].r);
		else q[i].ans=work1(q[i].l,q[i].r);
	}
	for(int i=1;i<=m;i++) printf("%lld\n",q[i].ans);
	return 0;
}