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>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define REP(i, n) rep(i, 0, n)
#define repb(i, a, b) for(int i = a; i >= b; i--)
#define all(a) a.begin(), a.end()
#define int long long
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
using namespace std;
typedef pair<int, int> P;
const int mod = 1000000007;
const int INF = 1e12;

int n, l, q;
int x[100010];
int d[100010];
int u[21][100010];

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    rep(i, 0, n) cin >> x[i];
    cin >> l >> q;
    int sum = 0, right = 0;
    for(int left = 0; left < n; left++){
        while(x[right] - x[left] <= l && right < n){
            right++;
        }
        d[left] = right - 1;
        // cout << left << " " << right << " " << d[left] << endl;
    }
    rep(i, 0, n) u[0][i] = d[i];
    rep(i, 0, 20){
        rep(j, 0, n){
            u[i + 1][j] = u[i][u[i][j]];
        }
    }
    rep(i, 0, q){
        int a, b;
        cin >> a >> b;
        a--; b--;
        if(a > b) swap(a, b);
        int now = a;
        int ans = 0;
        rep(k, 0, 30){
            // cout << now << endl;
            rep(j, 0, 20){
                if(u[j][now] >= b){
                    if(j != 0){
                        now = u[j - 1][now];
                        ans += pow(2, j - 1);
                    }else{
                        ans += 1;
                        k = 114514;
                    }
                    break;
                }
            }
        }
        cout << ans << endl;
    }
}