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>
using namespace std;

void f(vector<int> v) {
  for (int i = 0; i < v.size(); i++) {
    if (i == v.size() - 1)
    cout << v.at(i) << endl;
    else
    cout << v.at(i) << " ";
  }
}

int main() {
  int n; cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v.at(i);
  }
  if (prev_permutation(v.begin(), v.end()))
  f(v);
  next_permutation(v.begin(), v.end());
  f(v);
  if (next_permutation(v.begin(), v.end()))
  f(v);
}