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 PYTHON3):

Solution:
import itertools

if __name__ == '__main__':

	n = int(input())

	now = tuple(map(int,input().split()))

	seq = [i+1 for i in range(n)]
	seq2 = list(itertools.permutations(seq))

	ind = 0
	for j,k in enumerate(seq2):
		if k == now:
			ind = j

	if ind == 0:
		print(*seq2[ind])
		if len(seq2) != 1:
			print(*seq2[ind+1])
	elif ind+1 == len(seq2):
		print(*seq2[ind-1])
		print(*seq2[ind])
	else:
		print(*seq2[ind-1])
		print(*seq2[ind])
		print(*seq2[ind+1])