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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=100005;

int N,L,Q;
int x[MAXN],nxt[MAXN][25];
int ans[MAXN];
//int dis[MAXN];

int main()
{
	scanf("%d",&N);
	for(int i=1;i<=N;i++)
		scanf("%d",&x[i]);
	scanf("%d%d",&L,&Q);
	memset(nxt,0x3F,sizeof nxt);
	for(int i=1;i<=N;i++)
	{
		nxt[i][0]=lower_bound(x+1,x+N+1,x[i]+L)-x;
		if(x[nxt[i][0]]>x[i]+L)
			nxt[i][0]--;
	}
	for(int j=1;(1<<j)<N;j++)
		for(int i=1;i<=N;i++)
			if(nxt[i][j-1]!=0x3F3F3F3F&&nxt[nxt[i][j-1]][j-1]!=0x3F3F3F3F)
				nxt[i][j]=nxt[nxt[i][j-1]][j-1];

	for(int i=1,a,b;i<=Q;i++)
	{
		scanf("%d%d",&a,&b);
		if(a>b)
			swap(a,b);
		int ans=0;
		for(int j=20;j>=0;j--)
			if(nxt[a][j]<=b)
			{
				a=nxt[a][j];
				ans+=(1<<j);
			}
		if(a<b)
			ans++;
		printf("%d\n",ans);
	}

	return 0;
}