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>

int main() {
    int n;
    std::cin >> n;

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

    if (std::prev_permutation(A.begin(), A.end())) {
        for (auto it = A.begin(); it < A.end(); it++) {
            if (it != A.begin()) std::cout << " ";
            std::cout << *it;
        }
        std::cout << std::endl;
    }

    std::next_permutation(A.begin(), A.end());
    for (auto it = A.begin(); it < A.end(); it++) {
        if (it != A.begin()) std::cout << " ";
        std::cout << *it;
    }
    std::cout << std::endl;

    if (std::next_permutation(A.begin(), A.end())) {
        for (auto it = A.begin(); it < A.end(); it++) {
            if (it != A.begin()) std::cout << " ";
            std::cout << *it;
        }
        std::cout << std::endl;
    }
}