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 <algorithm>
#include <iostream>
#include <cstdio>
#include <map>
#include <numeric>
#include <cmath>
#include <set>
#include <sstream>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <complex>
#include <string.h>
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <iomanip>
#include <sys/time.h>
#include <random>
using namespace std;

#define endl '\n'
#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
#define UNIQ(v) (v).erase(unique((v).begin(), (v).end()), (v).end())

typedef long long ll;
typedef long double ld;
typedef pair<int, int> P;
typedef complex<double> comp;
typedef vector< vector<ld> > matrix;
struct pairhash {
public:
    template<typename T, typename U>
    size_t operator()(const pair<T, U> &x) const {
	size_t seed = hash<T>()(x.first);
	return hash<U>()(x.second) + 0x9e3779b9 + (seed<<6) + (seed>>2);
    }
};
const int inf = 1e9 + 9;
const ll mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1);

int n;
int x[100100];
int l, q;
int a[100100], b[100100];

int d[100100][20];

int calc(int idx0, int idx1) {
    int t = idx0;
    int res = 0;
    while (t < idx1) {
        int s = 19;
        while (d[t][s] > idx1) s--;
        t = d[t][s];
        res += (1<<s);
    }
    return res;
}

int calc0(int k) {
    int lb = k, ub = n+1;
    while (ub - lb > 1) {
        int mid = (ub + lb) / 2;
        if (x[mid] - x[k] <= l) {
            lb = mid;
        } else {
            ub = mid;
        }
    }
    return lb;
}

void solve() {
    for (int i = 0; i <= n; i++) {
        d[i][0] = calc0(i);
    }

    for (int i = 1; i < 20; i++) {
        for (int j = 0; j <= n; j++) {
            d[j][i] = d[d[j][i-1]][i-1];
        }
    }
    /*
    for (int j = 0; j < 5; j++) {
        for (int i = 0; i < n; i++) cerr << d[i][j] << " ";
        cerr << endl;
    }
    */
    for (int i = 0; i < q; i++) {
        if (a[i] > b[i]) swap(a[i], b[i]);
        cout << calc(a[i], b[i]) << endl;
    }
}

void input() {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> x[i];
    x[n] = x[n-1]+1;
    cin >> l >> q;
    for (int i = 0; i < q; i++) {
        cin >> a[i] >> b[i];
        a[i]--;
        b[i]--;
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(15);

    input();
    solve();
}