Problem: For given a sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$, print the previous permutation and the next permutation in lexicographic order.

Constraints

* $1 \leq n \leq 9$
* $a_i$ consist of $1, 2, ..., n$

Input

A sequence is given in the following format.


$n$
$a_0 \; a_1 \; ... \; a_{n-1}$


Output

Print the previous permutation, the given sequence and the next permutation in the 1st, 2nd and 3rd lines respectively. Separate adjacency elements by a space character. Note that if there is no permutation, print nothing in the corresponding line.

Examples

Input

3
2 1 3


Output

1 3 2
2 1 3
2 3 1


Input

3
3 2 1


Output

3 1 2
3 2 1

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
    int n;
    cin >> n;
    int arr[100];
    for (int i = 0; i < n; i++)
        cin >> arr[i];
    bool key;
    key = prev_permutation(arr, arr + n);
    if (key) {
        for (int i = 0; i < n; i++) {
            if (i == n - 1) {
                cout << arr[i] << endl;
            }
            else cout << arr[i] << " ";
        }
    }
    next_permutation(arr, arr + n);
    for (int i = 0; i < n; i++) {
        if (i == n - 1) {
            cout << arr[i] << endl;
        }
        else cout << arr[i] << " ";
    }
    key = next_permutation(arr, arr + n);
    if (key) {
        for (int i = 0; i < n; i++) {
            if (i == n - 1) {
                cout << arr[i] << endl;
            }
            else cout << arr[i] << " ";
        }
    }
}