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 a[10];

void print(int n)
{
    for(int i=0;i<n;i++)
        printf("%d%c",a[i],i==n-1?'\n':' ');
}

int main()
{
    int n;

    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }

    if(prev_permutation(a,a+n)) print(n);
    next_permutation(a,a+n);
    print(n);
    if(next_permutation(a,a+n)) print(n);

    return 0;
}