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 <bits/stdc++.h>  // clang-format off
using Int = long long;
#define REP_(i, a_, b_, a, b, ...) for (Int i = (a), lim##i = (b); i < lim##i; i++)
#define REP(i, ...) REP_(i, __VA_ARGS__, __VA_ARGS__, 0, __VA_ARGS__)
struct SetupIO { SetupIO() { std::cin.tie(nullptr), std::ios::sync_with_stdio(false), std::cout << std::fixed << std::setprecision(13); } } setup_io;
#ifndef dump
#define dump(...)
#endif  // clang-format on

/**
 *    author:  knshnb
 *    created: Sun May 31 16:53:00 JST 2020
 **/

void print(std::vector<Int>& a) {
    REP(i, a.size()) std::cout << a[i] << (i == a.size() - 1 ? "" : " ");
    std::cout << std::endl;
}
signed main() {
    Int n;
    std::cin >> n;
    std::vector<Int> a(n);
    REP(i, n) std::cin >> a[i];
    auto x = a, y = a;
    if (std::prev_permutation(x.begin(), x.end())) print(x);
    print(a);
    if (std::next_permutation(y.begin(), y.end())) print(y);
}