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<bits/stdc++.h>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) for(int i = 0; i < (b); i++)
#define all(a) (a).begin(), (a).end()
#define show(x)  cerr << #x << " = " << (x) << endl;

#define int long long

const int LOG = 20;
const int MAX_N = 100005;

const int INF = 1e8;
using namespace std;

int t[LOG][MAX_N];

signed main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n;
	cin >> n;

	vector<int> x(n);
	rep(i,n) cin >> x[i];

	rep(i,LOG) rep(j,MAX_N) t[i][j] = INF;

	int l, q;
	cin >> l >> q;

	//rep(i,n){ cerr << x[i] << ' '; } cerr << endl;

	rep(i,n){
		int right = n, left = 0;
		rep(j,LOG){
			int mid = (right + left ) / 2;
			if(x[mid] <= x[i] + l) left = mid;
			else right = mid;
		}

		t[0][i] = left;
	}

	rep(i,LOG - 1){
		rep(j,MAX_N){
			if(t[i][j] == INF) continue;
			t[i + 1][j] = t[i][ t[i][j] ];
		}
	}

	rep(i,q){
		int a, b;
		cin >> a >> b;
		a--; b--;
		if(a > b) swap(a,b);

		long long ans = 0;
		for(int i = LOG - 1; i >= 0; i--){
			if(t[i][a] < b){
				a = t[i][a];
				ans += (1LL << i);
			}
		}
		cout << ans + 1 << endl;
	}
}