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: Accepted)

#include<stdio.h>
#define N 100000

/*
How to test
% diff <(./a.out < b1in.txt) <(cat b1out.txt) 

  for( i=0 ; i < n ; i++){
    fprintf(stderr, "%d: C2[%d] = %c%c \n" ,__LINE__ ,i ,C2[i].suit ,C2[i].value );
  }

 */
void swap(short *a_p, short *b_p){
  int k;
  k    = *a_p;
  *a_p = *b_p;
  *b_p = k;
}


int prt(short* a_p, int p, int r ){
  int j;
  short x = a_p[r];
  int i = p-1;

  for( j = p ; j <= (r-1) ; j++){
    if(a_p[j]<=x){
      i++;
      swap(&a_p[i] ,&a_p[j]);
    }
  } 
  swap(&a_p[i+1] ,&a_p[r]);
  
  return i+1;
}

int main(){
  int n, i, j, k;
  int q;
  short x;
  short a[N];
  short a2[N][2] = {0};
  
  scanf("%d", &n);
  for ( i = 0; i <= n; i++ ) scanf("%d", &a[i]);

  x = a[n-1];
  q = prt(a,0,n-1);  

  for( i=(n-1) ; i>=0 ; i--){
    a2[i][0] = a[i];
    if(a[i]==x){
      a2[i][1] = 1; // piviot
      x = -1;
    }
  }


  for( i=0 ; i<n ; i++){
    if( a2[i][1] != 1){
      printf("%d" ,a2[i][0]);
    } else {
      printf("[%d]" ,a2[i][0]);
    }
    if( i != (n-1)){
      printf(" ");// Don't print more than enough space
    }
  }
  printf("\n");

  return 0; // Don't forget return 0
}