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 out(vector<int> a){
		bool isFirst = true;
		for(auto &&ai: a){
			if(!isFirst) cout<<" ";
			cout<<ai;
			isFirst &= false;
		}
		cout<<endl;;	
}

int main() {
	int n;cin>>n;
	vector<int> a(n);
	for(int i=0;i<n;++i)cin>>a[i];

	vector<int> prev(a), nx(a);
	prev_permutation(prev.begin(),prev.end());
	if(lexicographical_compare(prev.begin(),prev.end(),a.begin(),a.end())) out(prev);
	out(a);
	next_permutation(nx.begin(),nx.end());
	if(lexicographical_compare(a.begin(),a.end(),nx.begin(),nx.end())) out(nx);
	return 0;
}