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;
using ll = long long;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)

int main()
{
    ll n,output=3;
    cin >> n;
    vector<ll> a(n);
    rep(i,n) cin >> a[i];
    if(!prev_permutation(a.begin(), a.end())){
        next_permutation(a.begin(), a.end());
        output--;
    };


    rep(i,output){
        rep(i,n)
            cout << (i ? " " : "") << a[i];
        cout << endl;

        if(!next_permutation(a.begin() , a.end())) break;
    };

    return 0;
}