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)

#include<bits/stdc++.h>
using namespace std;

int main()
{
int N;scanf("%d",&N);
vector<int> x(N);vector<int> dist(N);
for(int i=0;i<N;i++){scanf("%d ",&x[i]);if(x[i]!=0)dist[i]=x[i]-x[i-1];}
int L,Q;scanf("%d\n%d",&L,&Q);int maxday=x[N-1]-x[0]+1;
vector<vector<int>> distday(maxday+1,vector<int>(N));
for(int i=0;i<N;i++){if(i!=N-1)distday[1][i]=i+1;distday[0][i]=i;}
int sum=0;bool check=true;int ie=0;
for(int i=0;i<N;i++)
{
  check=true;
  if(i>0)sum-=dist[i];
  while(check)
  {
    ie++;if(ie==N){ie--;break;}
    sum+=dist[ie];
    if(sum>L)
    {
      sum-=dist[ie];
      check=false;
      ie--;
    }
    distday[1][i]=ie;
  }
}
for(int a,b,i=0;i<Q;i++)
{
  scanf("%d %d",&a,&b);a--;b--;
  if(a>b)swap(a,b);
  int ed=a,days=0;
  while(ed<b)
  {
    days++;
    ed=distday[1][ed];
  }
  printf("%d\n",days);
}
}