Problem:
Partition

Quick sort is based on the Divide-and-conquer approach. In QuickSort(A, p, r), first, a procedure Partition(A, p, r) divides an array A[p..r] into two subarrays A[p..q-1] and A[q+1..r] such that each element of A[p..q-1] is less than or equal to A[q], which is, inturn, less than or equal to each element of A[q+1..r]. It also computes the index q.

In the conquer processes, the two subarrays A[p..q-1] and A[q+1..r] are sorted by recursive calls of QuickSort(A, p, q-1) and QuickSort(A, q+1, r).

Your task is to read a sequence A and perform the Partition based on the following pseudocode:

Partition(A, p, r)
1 x = A[r]
2 i = p-1
3 for j = p to r-1
4 do if A[j] <= x
5 then i = i+1
6 exchange A[i] and A[j]
7 exchange A[i+1] and A[r]
8 return i+1

Note that, in this algorithm, Partition always selects an element A[r] as a pivot element around which to partition the array A[p..r].

Input

The first line of the input includes an integer n, the number of elements in the sequence A.

In the second line, Ai (i = 1,2,...,n), elements of the sequence are given separated by space characters.

Output

Print the sorted sequence. Two contiguous elements of the sequence should be separated by a space character. The element which is selected as the pivot of the partition should be indicated by [  ].

Constraints

1 ≤ n ≤ 100,000

0 ≤ Ai ≤ 100,000

Sample Input 1

12
13 19 9 5 12 8 7 4 21 2 6 11

Sample Output 1

9 5 8 7 4 2 6 [11] 21 13 19 12

Solution in Ruby (judge verdict: Accepted)

def partition(a, p, r)
  x = a[r]
  i = p-1
  for j in p..r-1
    if a[j] <= x
      i += 1
      tmp = a[i]
      a[i] = a[j]
      a[j] = tmp
    end
  end
  tmp = a[i+1]
  a[i+1] = a[r]
  a[r] = tmp
  return i+1
 end


n = gets.chomp.to_i
input_array = gets.chomp.split.map(&:to_i)

q = partition(input_array, 0, n-1)
n.times do |i|
  if i == q
    print "[#{input_array[i]}]"
  else
    print input_array[i]
  end
  if i < (n-1)
    print " "
  end
end
print "\n"