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<vector>
#include<algorithm>
using namespace std;

int main(){

    int n;
    cin >> n;
    vector<int> vec(n);
    for(int i = 0; i < n; ++i){
        cin >> vec[i];
    }

    vector<int> origin = vec;

    if(prev_permutation(vec.begin(), vec.end())){
        for(auto it = vec.begin(); it != vec.end(); ++it){
            if(it != vec.begin()){
                cout << " ";
            }
            cout << *it;
        }
        cout << endl;
    }
    
    for(auto it = origin.begin(); it != origin.end(); ++it){
        if(it != origin.begin()){
            cout << " ";
        }
        cout << *it;
    }
    cout << endl;

    if(next_permutation(origin.begin(), origin.end())){
        for(auto it = origin.begin(); it != origin.end(); ++it){
            if(it != origin.begin()){
                cout << " ";
            }
            cout << *it;
        }
        cout << endl;
    }

    return 0;
}