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<cstdio>
#include<algorithm>
using namespace std;


long long N, x[100000], L;
int nx[100000][20];
int Q;


int main(){
    cin >> N;
    for(int i = 0; i < N; i++){
        cin >> x[i];
    }
    cin >> L;


    // dist [l, r] <= L
    int r = 0;
    for(int l = 0; l < N; l++){
        while(r+1 < N && x[r+1] - x[l] <= L){
            r++;
        }
        nx[l][0] = r;
    }
    for(int i = 0; i < N; i++){
        //cout << nx[i][0] << endl;
    }

    for(int k = 1; k < 20; k++){
        for(int i = 0; i < N; i++){
            nx[i][k] = nx[nx[i][k-1]][k-1];
        }
    }
    for(int k = 0; k < 20; k++){
        for(int i = 0; i < N; i++){
            //cout << nx[i][k] << " ";
        }
        //cout << endl;
    }
        

    
    cin >> Q;
    for(int i = 0; i < Q; i++){
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        if(a > b){
            int tmp = a;
            a = b;
            b = tmp;
        }
        int ans1, k1;
        if(a == b){
            cout << 0 << endl;
            continue;
        }
        // not begin, but end
        int begin = 0, end = 100001;
        while(end - begin > 1){
            int mid = (begin + end) / 2;
            int goal = a;
            for(int k = 0; k < 20; k++){
                if( (1 << k) & mid ){
                    goal = nx[goal][k];
                }
            }
            if(goal >= b){
                end = mid;
            }else{
                begin = mid;
            }
        }
        cout << end << endl;;

    }
}