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: Wrong Answer)

#include <bits/stdc++.h>
 
using namespace std;
 
// macro
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n-1;i >= 0;i--)
#define MOD 1000000007
#define INF 1000000007
#ifndef M_PI
    #define M_PI 3.14159265358979323846
#endif
#define ll long long
#define ull unsigned long long
#define FOR(i, m, n) for(ll i = m;i < n;i++)
#define SORT(ary) sort(begin(ary), end(ary))
#define VL vector<ll>
#define VVL vector<VL>
#define PL pair<ll, ll>
#define VC vector<char>
#define VVC vector<VC>
#define VI vector<int>
#define VVI vector<VI>

// def dp[i][j]
// position where can reach from i with 2^j times move
VVI dp(22, VI(pow(2, 20), 0));
VI x(pow(2, 20), INF);

int bSearch(int k, int l, int r) {
    if(r-l == 2) {
        if(x[(l+r)/2] > k) return l;
        else return (l+r)/2;
    }
    if(x[(l+r)/2] == k) return (l+r)/2;
    else if(x[(l+r)/2] > k) return bSearch(k, l, (l+r)/2);
    else return bSearch(k, (l+r)/2, r);
}

void initDp(int n, int l) {
    REP(i, n) {
        dp[0][i] = bSearch(x[i]+l, 0, pow(2, 20));
    }
    REP(i, 21) REP(j, pow(2, 20)) {
        dp[i+1][j] = dp[i][dp[i][j]];
    }
}

int count(int a, int b) {
    REPR(i, 21) {
        if(dp[i][a] < b) {
            return pow(2, i) + count(dp[i][a], b);
        }
    }
    return 1;
}

void solve() {
    int n;
    scanf("%d", &n);
    REP(i,n) {
        scanf("%d", &x[i]);
        printf("%d\n", x[i]);
    }
    int l,q;
    scanf("%d%d", &l, &q);
    initDp(n, l);

    // query
    REP(i, q) {
        int a,b;
        scanf("%d%d", &a, &b);
        // printf("%d\n", count(min(a, b)-1, max(a, b)-1));
    }
}

int main() {
  // init
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
}