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 C (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#define N 100000

int partition(int [], int, int);

int main()
{
  int i = 0,num,q,A[N];

  while(1)
   {
     scanf("%d",&num);
     if(num >= 1 && num <= N) break;
   }

  while(i<num)
   {
     scanf("%d",&A[i]);
     if(A[i] >= 1 && A[i] <= N) i++;
   }

  q = partition(A,0,num);

  for(i = 0;i < num-1;i++)
   {
     if(i == q) printf("[%d] ",A[i]);
     else printf("%d ",A[i]);
   }
  printf("%d\n",A[i]);

  return 0;
}

int partition(int a[], int p, int r)
{
  int i,j,x,swap;
  x = a[r-1];
  i = p-1;

  for(j = p;j < r;j++)
   {
     if(a[j] <= x)
      {
        i++;
        swap = a[i];
        a[i] = a[j];
        a[j] = swap;
      }
   }

  return i;
}