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 MAX 100000
int hairetu[MAX],a;
int partition(int,int);
int main(){
    int i,q;
    scanf("%d",&a);
    for(i=0;i<a;i++){
      scanf("%d",&hairetu[i]);
    }
    q=partition(0,a-1);
    for(i=0;i<a;i++){
      if(i)printf(" ");
      if(i==q)printf("[");
      printf("%d",hairetu[i]);
      if(i==q)printf("]");
    }
    printf("\n");
    return 0;
  }
int partition(int p,int r){
  int b,i,j,c;
  b=hairetu[r];
  i=p-1;
  for(j=p;j<r;j++){
    if(hairetu[j]<=b){
      i++;
      c=hairetu[i];
      hairetu[i]=hairetu[j];
      hairetu[j]=c;
    }
  }
  c=hairetu[i+1];
  hairetu[i+1]=hairetu[r];
  hairetu[r]=c;
  return i+1;
}