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>
#define int long long
#define r(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef tuple<int,int,string,int,string> T;

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