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<bits/stdc++.h>
using namespace std;

int dp[100005][20];
int x[100005];

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int a, b, n, L, Q;
	cin >> n;
	for(int i = 0; i < n; i ++) cin >> x[i];
	cin >> L >> Q;
	for(int i = 0; i < n; i ++) {
		int t = upper_bound(x, x + n, x[i] + L) - x;
		if(t == n) {
			if(x[i] + L > x[n - 1]) dp[i][0] = n;
			else dp[i][0] = n - 1;
		}
		else dp[i][0] = t - 1;
	}
	for(int i = 1; i < 20; i ++){
		for(int j = 0; j < n; j ++){
			if(dp[j][i - 1] == n){
				dp[j][i] = n;
			}
			else {
				dp[j][i] = dp[dp[j][i - 1]][i - 1];
			}
		}
	}
	for(int i = 0; i < Q; i ++){
		cin >> a >> b;
		a --; b --;
		if(x[a] > x[b]) swap(a, b);
		if(a == b){
			cout << 0 << '\n';
			continue;
		}
		int cur = a, ans = 0;
		for(int i = 19; i >= 0; i --){
			if(dp[cur][i] != n && x[dp[cur][i]] <= x[b] && cur != dp[cur][i]) {
				cur = dp[cur][i], ans += (1 << i);
			}
		}
		cout << ans << '\n';
	}
}