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;

typedef long long ll;
typedef pair<int,int> pii;

long long MOD = 1000000000 + 7;

int N;
int log_N;
vector<vector<int>> parent;
vector<ll> x;
ll L;

ll query(int a, int b){
  if(a == b){
    return 0;
  }

  int ret = 0;
  for(int k = log_N; k>= 0; k--){
    int next = parent[k][a];
    // cerr << "next: " << next << " a: " << a << " b: " << b << endl;

    if(next > b) continue;
    if(next == b) {
      ret += pow(2,k);
      break;
    } else {
      ret += pow(2,k);
      a = parent[k][a];
    }
  }
  return ret;
}

int main(){
  cout << setprecision(10);
  cin >> N;
  log_N = floor(log2(N));
  x.assign(N, 0);
  parent.assign(log_N + 1, vector<int>(N));

  for(int i = 0; i < N; i++){
    cin >> x[i];
  }
  cin >> L;
  cerr << "L: "  << L << endl;

  for(int i = 0; i < N; i++){
    // iから+Lするまでにあるところ最大
    auto itr = upper_bound(x.begin(), x.end(), x[i] + L);
    int index = itr - x.begin() - 1;
    // cerr << i << " " << index << endl;
    parent.at(0).at(i) = index;
  }
 
  for(int k = 0; k < log_N; k++){
    for(int i = 0; i < N; i++){
      if(parent[k][i] == -1){
        parent[k+1][i] = -1;
      } else { 
        parent[k+1][i] = parent[k][parent[k][i]];
        // cerr << k+1 << " " << i << " " << parent[k+1][i] << endl;
      }
    }
  }


  int Q;
  cin >> Q;
  for(int i = 0; i < Q; i++){
    int tmp1,tmp2;
    cin >> tmp1 >> tmp2;
    tmp1--;
    tmp2--;
    if(tmp1 > tmp2) swap(tmp1, tmp2);

    cout << query(tmp1, tmp2) << endl;
    
  }
}