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: Runtime Error)

#include <bits/stdc++.h>
#define int long long
using pii=std::pair<int,int>;
using namespace std;

const int maxn = 1e5 + 5, maxlogn = 18;

int n, l, q, x[maxn], a, b, r[maxlogn][maxlogn];
int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for(int i = 0; i < n; i++)
        cin >> x[i];
    cin >> l;
    for(int i = n - 1; i >= 0; i--)
    {
        int pos = upper_bound(x, x + n, x[i] + l) - x - 1;
        r[0][i] = pos;
    }
    for(int i = 1; i < maxlogn; i++)
        for(int j = n - 1; j >= 0; j--)
            r[i][j] = r[i - 1][r[i - 1][j]];
    cin >> q;
    for(int i = 0; i < q; i++)
    {
        cin >> a >> b;
        a--; b--;
        if(a > b)
            swap(a, b);
        int ans = 0;
        for(int j = maxlogn - 1; j >= 0; j--)
            if(r[j][a] < b)
            {
                a = r[j][a];
                ans += (1ll << j);
            }
        a = r[0][a];
        assert(a >= b);
        cout << ans + 1 << "\n";
    }
    return 0;
}