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, s, n) for (int i = s; i < n; i++)
#define ALL(a) a.begin(), a.end()
#define MOD 1000000007
using namespace std;
typedef long long ll;

int main() {
    int N; cin >> N;
    vector<ll> X(N);
    REP(i, 0, N) cin >> X[i];

    int L; cin >> L;
    int k = 1; while ((1 << k) < N) k++;
    int next[k][N] = {};
    REP(i, 0, N) {
        int idx = upper_bound(ALL(X), X[i] + L) - X.begin() - 1;
        next[0][i] = idx;
        // cout << X[idx] << endl;
    }
    REP(i, 0, k - 1) {
        REP(j, 0, N) {
            if (next[i][j] < 0) {
                next[i + 1][j] = -1;
            } else {
                next[i + 1][j] = next[i][next[i][j]];
            }
        }
    }

    int Q; cin >> Q;
    while (Q--) {
        int a, b; cin >> a >> b; a--; b--;
        if (a > b) swap(a, b);
        
        int ans = 0;
        for (int i = k - 1; i >= 0; i--) {
            if (next[i][a] < b) {
                ans += (1 << i);
                a = next[i][a];
            }
        }
        cout << ans + 1 << endl;
    }
    return 0;
}