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 <bits/stdc++.h>
using namespace std;

int main()
{
    vector<int> v;
    int n;
    cin >> n;

    while (n--)
    {
        int x;
        cin >> x;
        v.push_back(x);
    }
    int i;
    if (prev_permutation(v.begin(), v.end()))
        for (i = 0; i < v.size(); i++)
        {
            cout << v[i] << " \n"[i == v.size() - 1];
        }

    next_permutation(v.begin(), v.end());
    for (i = 0; i < v.size(); i++)
    {
        cout << v[i] << " \n"[i == v.size() - 1];
    }

    if (next_permutation(v.begin(), v.end()))
        for (i = 0; i < v.size(); i++)
        {
            cout << v[i] << " \n"[i == v.size() - 1];
        }
    return 0;
}