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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;

#define MAXI 100000

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

//input
int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    int hotels, maxdist, queries, temp, a, b;
    cin >> hotels;
    ordered_set loc;

    for(int one = 0; one < hotels; one++){
        cin >> temp;
        loc.insert(temp);
    }
    cin >> maxdist >> queries;

    //sparse table
    int temp2 = *loc.find_by_order(hotels-1);
    temp = log2(temp2);
    int st[hotels][temp+1]; //only for hotels that are not the leftmost

    //compute the furthest achievable hotel in 1 day
    for(int one = 0; one < hotels; one++){
        int temp = *loc.find_by_order(one);
        st[one][0] = loc.order_of_key(temp-maxdist);
    }

    //compute furthest hotel achievable in 2^temp days
    for(int one = 1; one < temp+1; one++){
        for(int two = 0; two < hotels; two++){
            st[two][one] = st[ (st[two][one-1]) ][one-1];
        }
    }
    //bug above

    //queries
    for(int one = 0; one < queries; one++){
        cin >> a >> b;
        a--; b--;
        if(a > b){swap(a,b);}
        int finans = 0;
        while(b > a){
            int days = 1, logdays = 0;
            while(st[b][logdays+1] > a){days*=2; logdays++;}
            finans += days;
            b = st[b][logdays];
        }
        cout << finans << endl;
    }
}