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 <stdio.h>
#include <algorithm>
#include <vector>
#define MAX 10
using namespace std;

void Print(vector<int> &vec)
{
    printf("%d", vec[0]);
    for (unsigned i = 1; i < vec.size(); i++)
    {
        printf(" %d", vec[i]);
    }
    printf("\n");
}

int main()
{
    int n;
    scanf("%d", &n);

    vector<int> vec(n), prevec, nextvec;
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &vec[i]);
    }

    prevec.assign(vec.begin(), vec.end());
    if (true == prev_permutation(prevec.begin(), prevec.end()))
    {
        Print(prevec);
    }

    Print(vec);
    nextvec.assign(vec.begin(), vec.end());
    if (true == next_permutation(nextvec.begin(), nextvec.end()))
    {
        Print(nextvec);
    }


    return 0;
}