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>

using namespace std;

const int N = 100005;

int x[N], p[N], d[N], pa[N][20];
int n, l, q, a, b;
vector<int> G[N];

void dfs(int v, int depth) {
	d[v] = depth;
	pa[v][0] = p[v];
	for(int k=1;k<20;k++)
		pa[v][k] = pa[pa[v][k-1]][k-1];
	for(int ch:G[v])
		dfs(ch, depth+1);
}

int main()
{
	cin >> n;
	for(int i=0;i<n;i++)
		cin >> x[i];
	cin >> l >> q;
	p[n-1]=n-1;
	for(int i=0;i<n-1;i++) {
		p[i] = upper_bound(x+i, x+n, x[i]+l)-x-1;
		G[p[i]].push_back(i);
	}
	dfs(n-1, 0);
	while(q--) {
		int ct = 0;
		cin >> a >> b;
		a--, b--;
		if(a > b) swap(a, b);
		for(int k=19;k>=0;--k){
			if(pa[a][k] < b)
				a = pa[a][k], ct += 1 << k;
		}
		cout << ++ct << endl;
	}
}