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

int main() {
	vector<int> vec; int an, set; bool check;
	cin >> an; vec.resize(an);
	for (int i = 0; i < an; i++) cin >> vec[i];
	check = false;
	for (int i = 0; i < an - 1; i++)
		if (vec[i] > vec[i + 1]) { check = true; break; }
	prev_permutation(vec.begin(), vec.end()); set = 0;
	if (check == true) {
		for (int i : vec) cout << (set++ ? " " : "") << i;
		cout << endl;
	}
	next_permutation(vec.begin(), vec.end()); set = 0;
	for (int i : vec) cout << (set++ ? " " : "") << i;
	cout << endl; check = false;
	for (int i = 0; i < an - 1; i++)
		if (vec[i] < vec[i + 1]) { check = true; break; }
	next_permutation(vec.begin(), vec.end()); set = 0;
	if (check == true) {
		for (int i : vec) cout << (set++ ? " " : "") << i;
		cout << endl;
	}
	return 0;
}