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>

#ifdef __LOCAL
    #define DBG(X) cout << #X << " = " << (X) << endl;
    #define SAY(X) cout << (X) << endl;
#else
    #define DBG(X)
    #define SAY(X)
#endif

#ifdef __LOCAL
    #include <filesystem>
    namespace fs = std::filesystem;
#endif

using namespace std;

using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

inline void fast_io() { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); };
template<typename T, typename S> inline ostream& operator<<(ostream& os, const pair<T, S> p) { cout << "[" << p.first << ";" << p.second << "]"; return os; }
template<typename T> inline ostream& operator<<(ostream& os, const vector<T>& v) { for (auto el : v) os << el << " "; return os; }
template<typename T> inline vector<T> fetch_vec(int sz) { vector<T> ret(sz); for (auto& elem : ret) cin >> elem; return ret; }

int N,L,Q;
vector<ll> X,A,B;


void input(){
  #ifdef __LOCAL
      fs::path p = __FILE__;
      fs::path input,output;
      input = output = p.parent_path();
      input += string("/input/") + string(p.stem()) + string(".txt");
      output += string("/output/") + string(p.stem()) + string(".txt");
      freopen(input.c_str(), "r", stdin);
      freopen(output.c_str(), "w", stdout);
  #endif

  fast_io();
  cin >> N;
  X = fetch_vec<ll>(N);
  cin >> L >> Q;
  for (int i = 0; i < Q; i++)
  {
    ll a,b;
    cin >> a >> b;
    a--,b--;
    if(a>b) swap(a,b);
    A.push_back(a);
    B.push_back(b);
  }
  

}



int solve(){
  int LogN = floor(log2(N));
  vector<vector<int>> next(LogN+1,vector<int>(N,0));
  for (int j = 0; j < N; j++)
  {
    auto itr = upper_bound(X.begin(),X.end(),L+X[j]);
    itr--;
    next[0][j] = itr-X.begin();
  }
  DBG(next[0])
  for (int k = 0; k < LogN; k++)
  {
    for (int i = 0; i < N; i++)
    {
      next[k+1][i] = next[k][next[k][i]];
    }
  }

  for (int i = 0; i < Q; i++)
  {
    int a=A[i],b=B[i];
    int ans=0;

    for(;;){
      int ng=-1,ok=LogN+1;
      while(ok-ng>1){
        int mid=(ng+ok)/2;
        if(next[mid][a]>=b) ok=mid;
        else ng=mid;
      }
      if(ng==-1){ans+=1; cout << ans << endl; break;}
      else{
        a = next[ng][a];
        ans += (1<<ng);
        }
    }

  }
  


  return 0;
}

int main()
{
  input();
  solve();
  return 0;
}