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>
#define fi first
#define se second
using namespace std;
typedef long long ll;
const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=1e9;
const int mod=1e9+7;
const int maxn=5e5+10;
int n,x[maxn];
int l,q,a,b;
int nex[maxn][20];
int main(){
    ios::sync_with_stdio(false);
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>x[i];
    }
    cin>>l>>q;
    for(int i=1;i<=n;i++) nex[i][0]=upper_bound(x+i,x+n+1,x[i]+l)-x-1;
    for(int i=0;i<19;i++){
        for(int j=1;j<=n;j++){
            nex[j][i+1]=nex[nex[j][i]][i];
        }
    }
    while(q--){
        cin>>a>>b;
        int ans=0;
        if(a>b) swap(a,b);
        for(int i=19;i>=0;i--){
            if(nex[a][i]<b){
                a=nex[a][i];
                ans+=(1<<i);
            }else if(nex[a][i]==b){
                ans+=(1<<i);
                break;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}