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 <bits/stdc++.h>
using namespace std;
using Pint = pair<int, int>;

int N, L, Q;
vector<int> x;
vector<Pint> q;

int main()
{
    cin >> N;
    x.resize(N);
    for (auto &val : x) cin >> val;
    x.push_back(x[N-1]+1);
    cin >> L >> Q;
    q.resize(Q);
    for (int i=0; i<Q; i++)
    {
        int t1, t2;
        cin >> t1 >> t2;
        
        if (t1>t2) swap(t1, t2);
        q[i] = Pint(t1-1, t2-1);
    }
    int lgN = 0;
    {
        int tmp = N;
        while (tmp)
        {
            lgN++;
            tmp /= 2;
        }
    }
    vector<vector<int> > next(lgN+1, vector<int>(N+1));
    for (int i=0; i<=N; i++)
    {
        next[1][i] = min((int)(lower_bound(x.begin(), x.end(), x[i]+L+1) - x.begin()) - 1, N);
    }

    for (int n=2; n<=lgN; n++)
    {
        for (int i=0; i<=N; i++)
        {
            next[n][i] = next[n-1][next[n-1][i]];
        }
    }

    for (auto query : q)
    {
        int ans = 0;
        int now = query.first;
        for (int n = lgN; n>0; n--)
        {
            if (next[n][now] < query.second)
            {
                ans += n;
                now = next[n][now];
            }
        }
        cout << ans + 1 << endl;
    }
}