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;
typedef long long ll;

int main(){
    ll n, t;
    vector<ll> v;
    cin >> n;
    for (auto i = 0; i< n; i++){
        cin >> t;
        v.push_back(t);
    }
    auto pv = v;
    if(prev_permutation(pv.begin(), pv.end())){
        for (auto i = pv.begin(); i != pv.end(); i++){
            cout << (i == pv.begin() ? "" : " ") << *i;
        }
        cout << endl;
    }
    for (auto i = v.begin(); i != v.end(); i++){
        cout << (i == v.begin() ? "" : " ") << *i;
    }
    cout << endl;
    auto nv = v;
    if(next_permutation(nv.begin(), nv.end())){
        for (auto i = nv.begin(); i != nv.end(); i++){
            cout << (i == nv.begin() ? "" : " ") << *i;
        }
        cout << endl;
    }
    return 0;
}