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 <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <deque>
#include <complex>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <ctime>
#include <iterator>
#include <bitset>
#include <numeric>
#include <list>
#include <iomanip>
#include <cassert>
#include <array>
#include <tuple>
#include <initializer_list>
#include <unordered_set>
#include <unordered_map>
#include <forward_list>
using namespace std;
using ll = long long;
#define rep(j,k) for (int j = 0; j < (int)(k); j++)

int n;
ll x[100010];
ll L;
int Q;
int a[100010], b[100010];

// x[i]から2^i回のジャンプで行けるところ
const int LOG = 33;
ll go[100010][LOG];

int solve(int a, int b) {
    int ans = 0;
    while (1) {
        if (go[a][0] >= b) return ans + 1;
        int i = 0;
        while (go[a][i] < b) ++i;
        --i;
        ans += 1 << i;
        a = go[a][i];
        // cout << a << ' ' << b << endl;
    }
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    while (cin >> n) {
        for (int i = 0; i < n; i++) {
            cin >> x[i];
        }
        x[n] = 2e9;
        cin >> L >> Q;
        for (int i = 0; i < n; i++) {
            int k = upper_bound(x + i, x + n, x[i] + L) - x;
            go[i][0] = k - 1;
        }
        for (int i = 0; i < LOG - 1; i++) {
            for (int j = 0; j < n; j++) {
                go[j][i + 1] = go[go[j][i]][i];
            }
        }
        for (int i = 0; i < Q; i++) {
            int a, b;
            cin >> a >> b;
            --a; --b;
            if (a > b) swap(a, b);
            //cout << 'a' << endl;
            cout << solve(a, b) << endl;
        }
    }
}