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>

int n, L, Q, lg;
std::vector <int> a;
std::vector <std::pair<int,int>> query;
std::vector <std::vector<int>> reach;

int main () {
	std::cin >> n;
	a.resize (n+1);
	for (int i = 1; i <= n; ++i)
		std::cin >> a[i];
	std::cin >> L >> Q;
	query.resize (Q);
	for (int i = 0; i < Q; ++i)
		std::cin >> query[i].first >> query[i].second;
	lg = std::__lg (n);
	if ((1<<lg) < n) ++lg;
	lg = 20;
	std::cerr << "lg = " << lg << "\n";
	reach.assign (n+1, std::vector<int> (lg+1, 0));
	for (int i = 1; i <= n; ++i) {
		int l = i, r = n, mid;
		while (l <= r) {
			mid = l + r >> 1;
			if (a[mid]-a[i] <= L)
				reach[i][0] = std::max(mid,reach[i][0]), l = mid+1;
			else
				r = mid-1;
		}
	}

	for (int j = 1; j <= lg; ++j)
		for (int i = 1; i <= n; ++i)
			reach[i][j] = reach[reach[i][j-1]][j-1];

	for (int i = 0; i < Q; ++i) {
		int go, st = query[i].first, ed = query[i].second;
		if (st > ed) std::swap (st, ed);
		for (go = 0; go <= lg; ++go)
			if (reach[st][go] >= ed) break;
		int ans = 0;
		for (; go >= 0 and st < ed; --go)
			if (reach[st][go] <= ed)
				st = reach[st][go], ans += (1<<go);
		std::cout << ans << "\n";
	}
	return 0;
}