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>
#define rep(i, n) for (int i = 0; i < (n); ++i) 
#define PI acos(-1)

using namespace std;
using ll = long long;
using P = pair<int, int>;
using LP = pair<ll, ll>;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    cin>>n;
    vector<int> as(n);
    rep(i, n) cin>>as[i];
    if(prev_permutation(as.begin(), as.end()))
        rep(i, n) cout<<as[i]<<(i != n - 1 ? ' ' : '\n');
    next_permutation(as.begin(), as.end());
    rep(i, n) cout<<as[i]<<(i != n - 1 ? ' ' : '\n');
    if(next_permutation(as.begin(), as.end()))
        rep(i, n) cout<<as[i]<<(i != n - 1 ? ' ' : '\n');
}