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

int main() {
	int	N;
	cin >> N;
	vector<int> vec(N);
	vector<int> temp(N);
	rep(i, N)
		cin >> vec.at(i);
	temp = vec;
	if (prev_permutation(vec.begin(), vec.end()))
	{
		rep(i, N)		
		{
			if (i)
				cout << ' ';
			cout << vec.at(i);
		}
		cout << endl;
	}
	rep(i, N)
	{
		if (i)
			cout << ' ';
		cout << temp.at(i);
	}
	cout << endl;
	if (next_permutation(temp.begin(), temp.end()))
	{
		rep(i, N)		
		{
			if (i)
				cout << ' ';
			cout << temp.at(i);
		}
		cout << endl;
	}
}