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>
typedef long long ll;
using namespace std;
#define all(x) (x).begin(), (x).end()
#define MOD 1000000007
int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    // dp[k][i]i番目のホテルから2^k日で到達可能な最右のホテル
    vector<vector<int>> dp(20, vector<int>(n, 0));
    for(size_t i = 0; i < n; i++) {
        cin >> a[i];
    }
    int l;
    cin >> l;
    for(size_t i = 0; i < n; i++) {
        auto itr = lower_bound(all(a), a[i] + l + 1) - a.begin();
        dp[0][i] = itr - 1;
    }
    for(int k = 0; k < 19; k++) {
        for(int i = 0; i < n; i++) {
            dp[k + 1][i] = dp[k][dp[k][i]];
        }
    }
    int q;
    cin >> q;
    while(q--) {
        int x, y;
        cin >> x >> y;
        x--;
        y--;
        if(x > y) {
            swap(x, y);
        }
        int day = 0;
        for(int i = 19; i >= 0; i--) {
            if(dp[i][x] < y) {
                x = dp[i][x];
                day += 1 << i;
            }
        }
        cout << day + 1 << endl;
    }
}