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 a[10],n,b[10];
  cin>>n;
  for(int i=0;i<n;i++){
    cin>>a[i];
    b[i]=a[i];
  }
  if(prev_permutation(a,a+n)==false) 1;
  else{
    for(int i=0;i<n;i++){
      if(i!=0) cout<<" ";
      cout<<a[i];
    }
    cout<<endl;
  }
  for(int i=0;i<n;i++){
    if(i!=0) cout<<" ";
    cout<<b[i];
  }
  cout<<endl;
  if(next_permutation(b,b+n)==true){
    for(int i=0;i<n;i++){
      if(i!=0) cout<<" ";
      cout<<b[i];
    }
    cout<<endl;
  }
  return 0;
}