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
from typing import List, Tuple


def print_elems(elems):
    print(" ".join([str(elem) for elem in elems]))


if __name__ == "__main__":
    n = int(input())
    nums = tuple(map(lambda x: int(x), input().split()))
    nums_permutated = list(itertools.permutations(sorted(nums)))

    if (1 == len(nums_permutated)):
        print(nums[0])
        exit(0)

    for idx in range(1, len(nums_permutated)):
        if (nums == nums_permutated[idx - 1]):
            print_elems(nums_permutated[idx - 1])
            print_elems(nums_permutated[idx])
            exit(0)
        if (nums == nums_permutated[idx]):
            print_elems(nums_permutated[idx - 1])
            print_elems(nums_permutated[idx])
            if (idx != len(nums_permutated) - 1):
                print_elems(nums_permutated[idx + 1])
            exit(0)