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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int n,x[100010],d[50][100010],i,j,a,b;
int main(){
  long long s,l;
  scanf("%d",&n);
  for(i=0;i<n;i++)scanf("%d",&x[i]);
  scanf("%lld",&l);
  for(i=j=s=0;i<n;i++){
    for(;j<n;j++){
      if(x[j]-x[i]>l)break;
    }
    d[0][i]=j-1;
  }
  //for(i=0;i<n;i++)printf("%d ",d[0][i]);printf("\n");
  for(i=0;i<40;i++){
    for(j=0;j<n;j++)d[i+1][j]=d[i][d[i][j]];
  }
  scanf("%d",&n);
  while(n--){
    scanf("%d %d",&a,&b);
    s=0;
    a--;
    b--;
    if(a>b){
      i=a;
      a=b;
      b=i;
    }
    for(j=40;j--;){//printf("%d %d\n",j,d[j][a]);
      if(d[j][a]<b){
	a=d[j][a];//printf("%d ",j);
	s+=1<<j;
      }
    }
    printf("%lld\n",s+1);
  }
  return 0;
}