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)

#pragma GCC optimize("Ofast")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <random>

using namespace std;
using namespace __gnu_pbds;

typedef
tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update> orderedSet;

#define pb push_back
#define F first
#define S second
#define all(a) (a).begin(), (a).end()

#define for1(i0, l0, r0) for (int i0 = l0; i0 < r0; ++i0)
#define for2(i0, l0, r0) for (int i0 = l0; i0 <= r0; ++i0)
#define forn(i0, n0) for (int i0 = 0; i0 < n0; ++i0)
#define forn1(i0, n0) for (int i0 = 1; i0 < n0; ++i0)
#define forr(i0, n0) for (int i0 = n0; i0 >= 0; --i0)
#define forr1(i0, r0, l0) for (int i0 = r0; i0 > l0; --i0)
#define forr2(i0, r0, l0) for (int i0 = r0; i0 >= l0; --i0)

#define Sort(a) sort(all(a))
#define Reverse(a) reverse(all(a))
#define relaxMax(a, b) a = max(a, b)
#define relaxMin(a, b) a = min(a, b)

typedef long long ll;
typedef long double ld;
#define ui unsigned int
#define ull unsigned long long

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef pair<string, string> pss;

typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<double> vd;
typedef vector<ld> vld;
typedef vector<bool> vb;
typedef vector<string> vs;

typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<pdd> vpdd;

typedef vector<vi> vvi;
typedef vector<vd> vvd;
typedef vector<vll> vvll;
typedef vector<vb> vvb;

#define vpss vector<pss>
#define vvs vector<vs>
#define vvpii vector<vpii>
#define vvpll vector<vpll>
#define vpt vector<pt>
#define vvvi vector<vvi>
#define vsi vector<set<int>>
#define ss second
#define ff first

#define printvi(arr) for (int x0 : arr) cout << x0 << ' '; cout << '\n';
#define printvll(arr) for (ll x0 : arr) cout << x0 << ' '; cout << '\n';
#define printpair(pair0) cout << pair0.F << ' ' << pair0.S << '\n';
#define printvp(arr) for (auto pair0 : arr) printpair(pair0);
#define initArray(arr, N0, X0) memset(arr, X0, N0 * sizeof(int))
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

void init() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    //cout.precision(16);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
}

const int LOG = 17;

void solve() {
    int n;
    cin >> n;
    vi x(n);
    forn(i, n) cin >> x[i];
    int l;
    cin >> l;
    vvi to(LOG, vi(n));
    int j = 0;
    forn(i, n) {
        while (j + 1 < n && x[j + 1] <= x[i] + l) j++;
        to[0][i] = j;
    }
    forn1(k, LOG) {
        forn(i, n) {
            to[k][i] = to[k - 1][to[k - 1][i]];
        }
    }
    int q;
    cin >> q;
    forn(i0, q){
        int a, b;
        cin >> a >> b;
        if (a > b) swap(a, b);
        a--, b--;
        int ans = 0;
        forr(i, LOG - 1){
            if (to[i][a] > b || (i > 0 && to[i - 1][a] == to[i][a])) continue;
            a = to[i][a];
            ans += 1 << i;
        }
        cout << ans << '\n';
    }
}

bool multitest = false;

int main() {
    init();
    int t = 1;
    if (multitest) cin >> t;
    forn(i, t) solve();
}