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 <string>
#include <algorithm>
#include <vector>

using namespace std;

int main() {
	int n;

	cin >> n;

	int A[n];

	for (int i = 0; i < n; i++) {
		cin >> A[i];
	}

	if (prev_permutation(A, A + n)) {
		for (int i = 0; i < n; i++) {
			if (i == 0)	cout << A[i];
			else cout << " " << A[i];
 		}
		cout << endl;
	}
	next_permutation(A, A + n);

	for (int i = 0; i < n; i++) {
		if (i == 0)	cout << A[i];
		else cout << " " << A[i];
	}
	cout << endl;

	if (next_permutation(A, A + n)) {
		for (int i = 0; i < n; i++) {
			if (i == 0)	cout << A[i];
			else cout << " " << A[i];
		}
		cout << endl;
	}
}