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 <iostream>
using namespace std;

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;
            int cpy = A[i];
            A[i] = A[j];
            A[j] = cpy;
        }
    }
    int cpy = A[r];
    A[r] = A[i + 1];
    A[i + 1] = cpy;
    
    return i + 1;
}

int main(void) {
    int A[100000];
    int n;
    
    cin >> n;
    
    for(int i = 0; i < n; ++i) {
        cin >> A[i];
    }
    
    int t = partition(A, 0, n - 1);
    
    for(int i = 0; i < t; ++i) {
        cout << A[i] << ' ';
    }
    cout << "[" << A[t] << "]";
    for(int i = t + 1; i < n; ++i) {
        cout << ' ' << A[i];
    }
    cout << endl;
    
    return 0;
}