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:
def solve():
    from sys import stdin
    f_i = stdin
    
    n = int(f_i.readline())
    A_s = f_i.readline().rstrip()
    
    if n == 1:
        print(A_s)
    else:
        A = tuple(map(int, A_s.split()))
        
        from itertools import permutations
        lex = tuple(permutations(range(1, n + 1), n))
        
        from bisect import bisect_left
        idx = bisect_left(lex, A)
        if idx == 0:
            print(A_s)
            print(' '.join(map(str, lex[idx+1])))
        elif idx == len(lex) - 1:
            print(' '.join(map(str, lex[idx-1])))
            print(A_s)
        else:
            print(' '.join(map(str, lex[idx-1])))
            print(A_s)
            print(' '.join(map(str, lex[idx+1])))

solve()