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: Runtime Error)

/// You just can't beat the person who never gives up
/// ICPC next year

#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include<bits/stdc++.h>
using namespace std ;
const int N = 1e5+5 ,sq = sqrt(N) ;

int n ,l ,q ,x[N] ,a ,b ,to[N] ;
vector<pair<int,int>> all ;
vector<pair<int,int>> ans[N] ;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;++i){
        scanf("%d",x+i);
        all.push_back({x[i],i});
    }
    scanf("%d",&l);
    sort(all.begin(),all.end());
    for(int i=0;i<all.size();++i) to[all[i].second] = i ;
    for(int i=0;i<all.size();++i){
        int k = min((int)all.size()-1 ,i+sq) ;
        int cur = i ;
        int t = 0 ;
        while(cur!=k){
            int st = cur ,en = k ;
            while(st<en){
                int mid = (st+en+1)/2 ;
                if(all[cur].first+l<all[mid].first) en = mid-1 ;
                else st = mid ;
            }
            ++t ;
            for(int j=cur+1;j<=st;++j) ans[i].push_back({j,t}) ;
            cur = st ;
        }
    }
    scanf("%d",&q);
    while(q--){
        scanf("%d%d",&a,&b);
        int aa = to[a] ;
        int bb = to[b] ;
        if(aa>bb) swap(aa,bb);
       // cout << aa << " " << bb << endl ;
        //for(auto go:ans[aa]) cout << go.first << " ";
       // cout << endl ;
        long long anss = 0 ;
        while(ans[aa].back().first<=bb){
            anss += ans[aa].back().second ;
            aa = ans[aa].back().first ;
        }
        if(aa!=bb) anss += ans[aa][bb-aa-1].second ;
        printf("%lld\n",anss);
    }
    return 0;
}