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>
using namespace std;
int main(){
	int i,j,last;
	int N,Q,L;
	int x[100001];
	int b[100001];
	int ans[100001];
	int c[100001];
	int d[100001];
		
	scanf("%d",&N);
	for(i=1;i<=N;i++){
		scanf("%d",&x[i]);
	}
	
	scanf("%d",&L);
	i=1;j=1;
	for(;i<=N;){
		for(;j<=N;){
			if((x[j]-x[i]<=L)){
				last = j;
				j++;
				
			}else{
				b[i]=last;
				i++;
			}
		}
		if(last==N){
			b[i]=last;
			i++;
		}
	}
	
	scanf("%d",&Q);
	for(i=1;i<=Q;i++){
		scanf("%d",&c[i]);
		scanf("%d",&d[i]);
	}
	for(i=1;i<=Q;i++){

		if(c[i]>d[i]){
			j=c[i];c[i]=d[i];d[i]=j;
		}

		if(L==1){
			ans[i]=d[i]-c[i];continue;
		}
		
		j=1;
		
		while(b[c[i]]<d[i]){
			c[i]=b[c[i]];
			j++;
		}
		ans[i]=j;
	}
	for(i=1;i<=Q;i++){
		printf("%d\n",ans[i]);
	}
	return 0;

	
}