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)

#define N 100005
#include <stdio.h>

int A[N];

void showArray(int A[], int n, int pos) {
    for (int i = 0; i < n; i++) {
        if (i == pos) {
            printf(" [%d]" + !i, A[i]);
        }
        else {
            printf(" %d" + !i, A[i]);
        }
    }
    printf("\n");
}

void swap(int *a, int *b) {
    int tmp = *a;
    *a = *b;
    *b = tmp;
}

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

int main() {
    int n = 0;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &A[i]);
    }
    int pos = partition(A, 0, n - 1);
    showArray(A, n, pos);
    return 0;
}