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: Accepted)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define in read()
#define re register
#define fur(i,a,b) for(re int i=a;i<=b;++i)
#define fdr(i,a,b) for(re int i=a;i>=b;--i)
#define cl(a,b) memset(a,b,sizeof(a))
#define jinitaimei signed
#define int long long
inline int read()
{
	int x=0;
	char ch=getchar();
	for(;!isalnum(ch);ch=getchar());
	for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
	return x;
}
const int xx=1e5+1612;
int to[xx][18];
int x[xx],n,L;
inline void init()
{
	n=in;
	fur(i,1,n) x[i]=in;
	L=in;
	int j=n;
	to[n][0]=j;
	fdr(i,n-1,1)
	{
		while(x[j]-x[i]>L) --j;
		to[i][0]=j;
	}
	fur(k,1,17) fur(i,1,n) to[i][k]=to[to[i][k-1]][k-1];
}
inline int power(int b,int k)
{
	int res=1;
	for(;k;k>>=1,b=b*b) if(k&1) res=res*b;
	return res;
}
inline void sol(int a,int b)
{
	if(a>b) swap(a,b);
	int j=a,g=17;
	int ans=0;
	while(j<b)
	{
		while(to[j][g]>b&&g) --g;
		if(b==n) while(to[j][g]==b&&g) --g;
		j=to[j][g];
		ans+=power(2,g);
	}
	printf("%lld\n",ans);
}
jinitaimei main()
{
	init();
	int T=in;
	while(T--) sol(in,in);
	return 0;
}