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() 
{
    int n, m, i;
    scanf("%d", &n);
    
    vector<int> v1;
    vector<int> v2;
    
    for(i = 0; i < n; i++)
    {
        scanf("%d", &m);
        v1.emplace_back(m);
        v2.emplace_back(m);
    }
    
    if(prev_permutation(v1.begin(), v1.end()))
    {
        for(i = 0; i < n-1; i++)
            printf("%d ", v1[i]);
        printf("%d\n", v1[i]);
    }

    for(i = 0; i < n-1; i++)
        printf("%d ", v2[i]);
    printf("%d\n", v2[i]);
    
    if(next_permutation(v2.begin(), v2.end()))
    {
        for(i = 0; i < n-1; i++)
            printf("%d ", v2[i]);
        printf("%d\n", v2[i]);
    }
    
    return 0;
}