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>
int patition(int a[100000], int p, int r) {
    int x, i, j, tmp;
 
    x = a[r];
    i = p - 1;
    for (j = p; j < r; j++) {
        if (a[j] <= x) {
            i += 1;
            tmp = a[j];
            a[j] = a[i];
            a[i] = tmp;
        }
    }
    tmp = a[i+1];
    a[i+1] = a[r];
    a[r] = tmp;
    return i + 1;
}
 
int main(void) {
    int in[100000], n, i, p;
    const char *fmt = "%d";
 
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
        scanf("%d", in+i);
    }
    p = patition(in, 0, n-1);
 
    for (i = 0; i < p; i++) {
        printf(fmt, in[i]);
        fmt = " %d";
    }
    printf(" [%d]", in[p]);
    for (i = p+1; i < n; i++) {
        printf(fmt, in[i]);
    }
    putchar('\n');
    return 0;
}