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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

struct Doubling
{
    const int LOG;
    vector< vector< int > > table;
    
    Doubling(int sz, int64_t lim_t) : LOG(64 - __builtin_clzll(lim_t))
    {
        table.assign(LOG, vector< int >(sz, -1));
    }

    void set_next(int k, int x)
    {
        table[0][k] = x;
    }

    void build()
    {
        for(int k = 0; k + 1 < LOG; k++) {
            for(int i = 0; i < table[k].size(); i++) {
                if(table[k][i] == -1) table[k + 1][i] = -1;
                else table[k + 1][i] = table[k][table[k][i]];
            }
        }
    }

    int query(int k, int64_t t)
    {
        for(int i = LOG - 1; i >= 0; i--) {
            if((t >> i) & 1) k = table[i][k];
        }
        return k;
    }
};


int main(void) {
    int N;
    cin >> N;
    vector<int> x(N);
    for(int i=0; i<N; i++) {
        cin >> x[i];
    }
    
    int L;
    cin >> L;
    vector<int> l(N), r(N);
    for(int i=0; i<N; i++) {
        l[i] = distance(x.begin(), lower_bound(x.begin(), x.end(), x[i] - L));
        r[i] = distance(x.begin(), upper_bound(x.begin(), x.end(), x[i] + L)) - 1;
        // cerr << l[i] << ' ' << r[i] << endl;
    }

    Doubling left(N, 1000000), right(N, 1000000);
    for(int i=0; i<N; i++) {
        left.set_next(i, l[i]);
        right.set_next(i, r[i]);
    }
    left.build();
    right.build();

    int Q;
    cin >> Q;
    while(Q--) {
        int a, b;
        cin >> a >> b;
        --a; --b;
        if(a < b) {
            int ok = 100000;
            int ng = 0;
            while(ok > ng + 1) {
                int mid = (ok + ng) / 2;
                if(right.query(a, mid) >= b) {
                    ok = mid;
                } else {
                    ng = mid;
                }
            }
            cout << ok << endl;
        } else {
            int ok = 100000;
            int ng = 0;
            while(ok > ng + 1) {
                int mid = (ok + ng) / 2;
                if(left.query(a, mid) <= b) {
                    ok = mid;
                } else {
                    ng = mid;
                }
            }
            cout << ok << endl;
        }
    }
    return 0;
}