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: Time Limit Exceeded)

#include<bits/stdc++.h>

#define INF 1e9
#define llINF 1e18
#define MOD 1000000007
#define pb push_back
#define mp make_pair 
#define F first
#define S second
#define ll long long
#define ALL(a) (a).begin(),(a).end()
#define Yes(hoge) cout<<((hoge)?"Yes":"No")<<endl;
#define YES(hoge) cout<<((hoge)?"YES":"NO")<<endl;
typedef struct aaaaa{
  int x,y,t;
}Grid;
using namespace std;
int main(){
  int n;cin>>n;
  vector<ll>ori(n);
  vector<vector<ll> >vvll(n);
  map<ll,int>mli;
  bool used[n+10]={};
  for(int i=0;i<n;i++)
    cin>>ori[i];
  ll L;cin>>L;
  int q;cin>>q;
  ori.pb(llINF);
  int cnt=0;
  for(int i=0;i<n;i++){
    if(used[i])continue;
    vvll[cnt].pb(ori[i]);
    used[i]=true;
    ll hoge=ori[i];
    mli[i]=cnt;
    while(hoge!=llINF){
      auto po = upper_bound(ALL(ori),hoge+L);
      if(*po == llINF){
	if(ori[ori.size()-2]-hoge<=L){
	  mli[ori.size()-2]=cnt;
	  used[ori.size()-2]=true;
	  vvll[cnt].pb(ori[ori.size()-2]);
	  break;
	}
      }
      int num=(po-ori.begin())-1;
      if(num<0)num=0;
      used[num]=true;
      hoge = ori[num];
      mli[num]=cnt;
      vvll[cnt].pb(ori[num]);
    }
    cnt++;
  }
  /* for(int i=0;i<vvll.size();i++){
    for(int j=0;j<vvll[i].size();j++)
      cout<<vvll[i][j]<<" ";
    cout<<endl;
    }*/

  for(int i=0;i<q;i++){
    ll le,ri;cin>>le>>ri;
    //cout<<mli[min(le,ri)-1]<<endl;
    //cout<<ori[min(le,ri)-1]<<" "<<ori[max(le,ri)-1]<<endl;
    ll num=min(le,ri)-1;
    int hhh=mli[num];
    auto a = lower_bound(ALL(vvll[hhh]),ori[min(le,ri)-1]);
    auto b = lower_bound(ALL(vvll[hhh]),ori[max(le,ri)-1]);
    printf("%d\n",(b-a));
  }
  return 0;
}