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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <algorithm>
#include <cctype>
#include <cmath>
#include <queue>
#include <map>
#include <numeric>
#include <unordered_map>
#include <iomanip>
#include <functional>
#include <bitset>
#include <complex>

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define repi(i,a,b) for(int i=int(a);i<int(b);i++)
#define all(x) (x).begin(),(x).end()

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

typedef long long ll;

using namespace std;


int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin>>n;
  vector<int>city(n);
  rep(i,n)cin>>city[i];
  int l,q;
  cin>>l>>q;
  vector<vector<int>>db(n,vector<int>(20,n));
  int r=0;
  rep(i,n)
  {
    while(r+1<n&&city[r+1]<=city[i]+l)r++;
    db[i][0]=r;
  }
  rep(i,19)
  {
    rep(j,n)
    {
      db[j][i+1]=db[db[j][i]][i];
    }
  }
  rep(i,q)
  {
    int a,b;
    cin>>a>>b;
    a--;b--;
    if(a>b)swap(a,b);
    int cnt=0;
    while(1)
    {
      auto itr=lower_bound(all(db[a]),b);
      if(itr==db[a].begin()){cnt++;break;}
      cnt+=(1<<(itr-db[a].begin()-1));
      a=*(itr-1);
    }
    cout<<cnt<<"\n";

  }
  cout<<flush;


  return 0;

}