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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define rep(i, x) for(ll i = 0; i < x; i++)
#define all(a) (a).begin(),(a).end()
using ll = long long;
using ld = long double;
using namespace std;
const ll INF = 1000000000000000000;
const ll mod = 1000000007;
const ld pi = 3.141592653589793238;
//printf("%.10f\n", n);

ll gcd(ll a, ll b) {
	if (a < b)swap(a, b);
	if (a % b == 0)return b;
	return gcd(b, a % b);
}
ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}
ll modpow(ll x, ll y) {
	ll res = 1;
	while (y) {
		if (y % 2) { res *= x; res %= mod; }
		x = x * x % mod; y /= 2;
	}
	return res;
}

ll G[12345678], test[12345678];
signed main() {
	ll n, l, q, a, b; cin >> n;
	for (int h = 0; h < n; h++) { cin >> test[h]; }
	cin >> l >> q;
	ll mae = -1, ato = 0;
	while (true) {
		if (mae == n - 1) { break; }
		mae++;
		while (true) {
			if (ato != n - 1 && test[ato + 1] - test[mae] <= l) { ato++; }
			else { break; }
		}
		//cout << ' ' << mae + 1 << ' ' << ato + 1 << endl;
		G[mae + 1] = ato + 1;
	}
	vector<vector<ll>> v(n + 5); vector<ll> oya(n + 1), visit(n + 1);
	for (int h = 1; h < n; h++) {
		if (visit[h] == 0) {
			ll cnt = h; visit[cnt] = 1; oya[cnt] = h;
			v[h].push_back(cnt);
			while (true) {
				//cout << cnt << ' ' << G[cnt] << endl;
				cnt = G[cnt]; visit[cnt] = 1; oya[cnt] = h;
				//v[h].push_back(cnt);
				//cout << h << ' ' << cnt << endl;
				if (cnt == n) { break; }
			}
		}
	}
	for (int h = 0; h < q; h++) {
		cin >> a >> b; if (a > b) { swap(a, b); }
		ll c = lower_bound(v[oya[a]].begin(), v[oya[a]].end(), a) - v[oya[a]].begin();
		ll d = lower_bound(v[oya[a]].begin(), v[oya[a]].end(), b) - v[oya[a]].begin();
		cout << d - c << endl;
	}
	return 0;
}