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<bitset>
#include<string>
#include<cmath>
#include<algorithm>
#include<stack>
#include<list>
#include<queue>
#include<map>
using namespace std;
#define rep(i,x) for(int i=0;i<x;i++)
#define re(i,x,y) for(int i=x;i<y;i++)
int INF=1e9;
const long long mod=1e9+7;
typedef long long ll;

int main(){
  int n;
  cin>>n;
  vector<int>a(n);
  rep(i,n){
    cin>>a[i];
  }
  if(prev_permutation(a.begin(),a.end())){
    rep(i,n){
      if(i)cout<<' ';
      cout<<a[i];
    }
    cout<<endl;
  }
  next_permutation(a.begin(),a.end());
  rep(i,n){
    if(i)cout<<' ';
    cout<<a[i];
  }
  cout<<endl;
  if(next_permutation(a.begin(),a.end())){
    rep(i,n){
      if(i)cout<<' ';
      cout<<a[i];
    }
    cout<<endl;
  }
}