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;
#define fs first
#define sc second
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<LL,LL> P;
const LL mod=1e9+7;
const LL LINF=1LL<<62;

LL dp[31][100000];
map<int,LL> ma;

LL dfs(int a,int b,LL sum){
    for(int k=0;k<31;k++){
        if(b==dp[k][a]){
            return sum+(1<<k);
        }
        if(b<dp[k][a]){
            if(k==0) return sum+1;
            return dfs(dp[k-1][a],b,sum+(1<<k));
        }
    }
    return 0;
}

int main(){
    int N;
    cin >> N;
    vector<LL> X;
    for(int i=0;i<N;i++){
        LL x;
        cin >> x;
        X.eb(-x);
        ma[i]=x;
    }
    sort(ALL(X));
    LL L,Q;
    cin >> L >> Q;
    for(int i=0;i<N;i++){
        dp[0][i]=X.end()-lower_bound(X.begin(),X.end(),-(ma[i]+L))-1;
    }
    for(int k=1;k<31;k++){
        for(int i=0;i<N;i++){
            dp[k][i]=dp[k-1][dp[k-1][i]];
        }
    }
    while(Q--){
        int a,b;
        cin >> a >> b;
        a--,b--;
        if(a>b) swap(a,b);
        cout << dfs(a,b,0) << endl;
    }
    return 0;
}