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;
int main(){
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++){
    cin >> a[i];
  }
  bool b;
  b = prev_permutation(a.begin(), a.end());
  if (b){
    for (int i = 0; i < n; i++){
      cout << a[i];
      if (i < n - 1){
        cout << ' ';
      }
    }
    cout << endl;
  }
  next_permutation(a.begin(), a.end());
  for (int i = 0; i < n; i++){
    cout << a[i];
    if (i < n - 1){
      cout << ' ';
    }
  }
  cout << endl;
  b = next_permutation(a.begin(), a.end());
  if (b){
    for (int i = 0; i < n; i++){
      cout << a[i];
      if (i < n - 1){
        cout << ' ';
      }
    }
    cout << endl;
  }
}