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>
typedef long long ll;
typedef long double ld;
const int INF=1e9,MOD=1e9+7,ohara=1e6;
const ll LINF=1e18;
using namespace std;

#define rep(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define rrep(i,a,b) for(int i=(a);i<(b);i++)
#define rrrep(i,a,b) for(int i=(a);i>=(b);i--)
#define all(v) (v).begin(), (v).end()
#define Size(n) (n).size()
#define Cout(x) cout<<(x)<<endl

ll n,cnt,ans,a,b,c,d,tmp,tmpp,m,h,w,x[ohara],y,sum,pos,k,q,l,r[35][ohara];
ld doua;
int dy[]={1,0,-1,0};
int dx[]={0,1,0,-1};
//int dy[]={-1,0,1,-1,1,-1,0,1};
//int dx[]={-1,-1,-1,0,0,1,1,1};
string alph("abcdefghijklmnopqrstuvwxyz"),s;
bool fl;
struct edge{int to,cost;};
unordered_map<ll,ll> mp;

//-------------------------↓↓↓↓↓↓------------------------

int main(void){
       cin.tie(0);
    ios::sync_with_stdio(false);

    cin>>n;
    set<ll> st;
    rep(i,n)cin>>x[i],st.insert(x[i]),mp[x[i]]=i;
    cin>>l>>q;
    rep(i,n){
        auto iti=st.upper_bound(x[i]+l);
        iti--;
        pos=*iti;
        pos=mp[pos];
        r[0][i]=pos;
    }
    rrep(i,1,31){
        rep(j,n){
            r[i][j]=r[i-1][r[i-1][j]];
        }
    }
    rep(i,q){
        ll fr,to,now;
        cin>>fr>>to;
        fr--;to--;
        if(fr>to){
            tmp=fr;
            fr=to;
            to=tmp;
        }
        now=fr;
        rrrep(j,30,0){
            if(r[j][now]<=to){
                now=r[j][now];
                ans+=1<<j;
            }
        }
        Cout(ans);
        ans=0;
    }
    return 0;
}