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 Ruby (2.3.3) (judge verdict: Wrong Answer)

n=gets.to_i
bd=0
hotel=[]
hotel << 0 #dummy
gets.split.each do |x|
  t=x.to_i
  hotel << t
  bd |= 1 << t
end
#p bd.to_s 2
L=gets.to_i
l=(1 << L+2)-1

cnt=0
#p L
#p (l << L -1).to_s(2)
#puts
(gets.to_i).times do
  a,b=gets.split.map &:to_i
  if a>b then a^=b;b^=a;a^=b end
  path = ( bd  & ( ( 1 << hotel[b] +1) -1) ) >> hotel[a] 
#  p path.to_s 2

  while path > 1 do
    
    path >>= Math.log2(path & l).floor
#    p path.to_s 2
 
#    p Math.log2(path & l)
#    path >>= Math.log2(path & l).floor
    #print path.to_s(2)," ", Math.log2(path & l).ceil,"\n"
  #  path=0
    cnt+=1
  end   
#  end
  p cnt
  cnt=0
end