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

int main() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (int i = 0;i < n;++i) {
        cin >> v[i];
    }
    vector<int> p(v), nx(v);
    prev_permutation(p.begin(), p.end());
    next_permutation(nx.begin(), nx.end());
    if (p < v){
        for(int i = 0;i < n;++i){
            if(i){
                cout << " ";
            }
            cout << p[i];
        }
        cout << endl;
    }
    for(int i = 0;i < n;++i){
        if(i){
            cout << " ";
        }
        cout << v[i];
    }
    cout << endl;
    if (nx > v){
        for(int i = 0;i < n;++i){
            if(i){
                cout << " ";
            }
            cout << nx[i];
        }
        cout << endl;
    }
}