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 a[10000];
    for (int i = 0; i < n; i++)
    {
        int m;
        cin >> m;
        a[i] = m;
    }
   if( prev_permutation(a, a + n)!=false)
    for (int i = 0; i < n; i++)
    {
        cout << a[i];
        if (i == n - 1)cout << endl;
        if (i != n - 1)cout << " ";
    }

   next_permutation(a, a + n);
        for (int i = 0; i < n; i++)
        {
            cout << a[i];
            if (i == n - 1)cout << endl;
            if (i != n - 1)cout << " ";
       }
       if( next_permutation(a, a + n)!=false)
        for (int i = 0; i < n; i++)
        {
            cout << a[i];
            if (i == n - 1)cout << endl;
            if (i != n - 1)cout << " ";
        }
}