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: Accepted)

#include "bits/stdc++.h"
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vvi = vector<vi>;
using vvl = vector<vl>;
const ll INF = 1LL << 60;
const ll MOD = 1000000007;
template <class T>
bool chmax(T &a, const T &b) {
    return (a < b) ? (a = b, 1) : 0;
}
template <class T>
bool chmin(T &a, const T &b) {
    return (b < a) ? (a = b, 1) : 0;
}
template <class C>
void print(const C &c, std::ostream &os = std::cout) {
    std::copy(std::begin(c), std::end(c), std::ostream_iterator<typename C::value_type>(os, " "));
    os << std::endl;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vl x(n);
    for (int i = 0; i < n; ++i) {
        cin >> x[i];
    }
    ll l, q;
    cin >> l >> q;
    vl a(q), b(q);
    for (int i = 0; i < q; ++i) {
        cin >> a[i] >> b[i];
        a[i]--, b[i]--;
    }
    vvl dbl(n, vl(20, n - 1));
    int nxt = 0;
    for (int i = 0; i < n; ++i) {
        while (x[nxt] - x[i] <= l && nxt < n) {
            nxt++;
        }
        nxt--;
        dbl[i][0] = nxt;
    }
    for (int i = 1; i <= 17; ++i) {
        for (int j = 0; j < n; ++j) {
            dbl[j][i] = dbl[dbl[j][i - 1]][i - 1];
        }
    }
    auto check = [&](int mid, int i) {
        int l = min(a[i], b[i]);
        int r = max(a[i], b[i]);
        for (int j = 17; j >= 0; --j) {
            if (mid & 1 << j)
                l = dbl[l][j];
        }
        return r <= l;
    };
    for (int i = 0; i < q; ++i) {
        int ok = n;
        int ng = 0;
        while (ng + 1 != ok) {
            int mid = (ok + ng) / 2;
            if (check(mid, i))
                ok = mid;
            else
                ng = mid;
        }
        cout << ok << "\n";
    }
    return 0;
}