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<vector>
#include<algorithm>
using namespace std;

int main() {
	int n;cin>>n;
	vector<int> x(n);
	for(int i=0;i<n;i++){
		cin>>x[i];
//		cout<<x[i];
	}
	int l,q;cin>>l>>q;
	vector<int> a(q),b(q);
	for(int i=0;i<q;i++){
		cin>>a[i]>>b[i];
		a[i]--;
		b[i]--;
	}
	vector<vector<int>> r,r2;
	vector<int> temp(n),ex(n);
	int j=1;
	for(int i=0;i<n-1;i++){
		while((x[i]+l>=x[j])*(j<n)){
			j++;
		}
		temp[i]=j-1;
	}
	temp[n-1]=n-1;
	r.push_back(temp);
	int flag=1;
	int count=1;
	while(flag){
		flag=0;
		ex=temp;
		temp=vector<int>(n,0);
//		for(int i=0;i<ex.size();i++){
//			cout<<ex[i];
//		}
		for(int i=0;i<n;i++){
			temp[i]=ex[ex[i]];
			if(temp[i]!=n-1){
				flag=1;
			}
		}
		r.push_back(temp);
		count++;
	}
	count--;
	r2=vector<vector<int>>(n,vector<int>(count,0));
	for(int i=0;i<n;i++){
		for(int j=0;j<count;j++){
			r2[i][j]=r[j][i];
		}
	}
//	for(int i=0;i<r.size();i++){
//		for(int j=0;j<r[i].size();j++){
//			cout<<r[i][j]<<' ';
//		}
//		cout<<endl;
//	}
	vector<int> sp(count);
	sp[0]=1;
	for(int i=1;i<count;i++){
		sp[i]=sp[i-1]*2;
	}
	for(int i=0;i<q;i++){
		int ans=0;
		int start=a[i];
		int goal=b[i];
		if(a[i]>b[i]){
			start=b[i];
			goal=a[i];
		}
		while(1){
			int add=upper_bound(r2[start].begin(),r2[start].end(),goal)-r2[start].begin();
//			cout<<add;
			add--;
			ans+=sp[add];
			start=r2[start][add];
			if(start==goal)break;

		}
		cout<<ans<<endl;
	}
	return 0;
}