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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
int x[100005]={},dp[100005][20],lg[20]={1}; //倍增
int main()
{
	int n,m,q,a,b,temp,ans,d,last,k;
	scanf("%d",&n); last=n;
	for(int i=1;i<=n;i++) scanf("%d",&x[i]);
	scanf("%d %d",&m,&q);
	for(int i=1;i<20;i++) lg[i]=lg[i-1]*2; //准备好2的幂
	for(int i=n;dp[1][0]==0;i--){ //预处理走一步的情况
		d=x[i]-m; //极限距离
		if(x[last]<d) continue; //本次结果和上一次一样 没必要搜索
		for(int j=last;j>=1&&x[j]>=d;j--,last--)
			dp[j][0]=i;
	}
	for(int j=1;j<=16;j++) //一次走65536步是可能的 走2^17是不可能的
		for(int i=1;i<=n;i++)
			dp[i][j]=dp[dp[i][j-1]][j-1]; //核心算法：在i点走2^j步等效于在i点走2^(j-1)步后再走2^(j-1)步！
	while(q>0){
		q--; ans=0;
		scanf("%d %d",&a,&b);
		if(a>b){ //a到b和b到a等效
			temp=a;
			a=b;
			b=temp;
		}
		while(dp[a][0]<b){
			for(k=0;k<=16;k++) //大步流星
				if(dp[a][k]>=b)
					break;
			ans+=lg[k-1];
			a=dp[a][k-1];
		}
		printf("%d\n",ans+1);
	}
	return 0;
}