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<iostream>
#include<vector>
#include<map>
#include<algorithm>
#include<string>
#include<stdio.h>
#include<stdlib.h>
#include<set>
#include<cmath>
#include<queue>
#define MAX 1000050
#define ll long long int
using namespace std;

int n,l;
int q;
vector<int>x;
int t;
int a ,b;
int main(){
  scanf("%d",&n);
  for(int i=0;i<n;i++){
    scanf("%d",&t);
    x.push_back(t);
  }
  scanf("%d", &l);
  scanf("%d",&q);
  for(int i=0;i<q;i++){
    scanf("%d %d",&a,&b);
    int cur = 0;
    int cnt = 0;
    if(a > b) swap(a,b);
    int start = a-1;
    for(int j=a;j<=b-1;j++){
      if(abs(x[start]-x[j]) <= l){
        cur = max(cur,abs(x[start]-x[j]));
      }
      if(((j+1)<=b-1) && (cur+abs(x[j]-x[j+1]))>=l){
        cur = 0;
        cnt++;
        start = j;
      }
      if(j==b-1 && cur>0) {
        cnt++;
      }
    }
    printf("%d\n",cnt);
  }
  return 0;
}