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 a[1010];
int main()
{
	int n;
	scanf("%d",&n);
	int i;
	for(i=0; i<n; i++) scanf("%d",&a[i]);
	if(prev_permutation(a, a+n))
	{
		for(i=0; i<n; i++)
		{
			if(i == n-1) printf("%d\n",a[i]);
			else printf("%d ",a[i]);
		}
	}
	next_permutation(a, a+n);
	for(i=0; i<n; i++)
	{
		if(i == n-1) printf("%d\n",a[i]);
		else printf("%d ",a[i]);
	}
	if(next_permutation(a, a+n))
	{
		for(i=0; i<n; i++)
		{
			if(i == n-1) printf("%d\n",a[i]);
			else printf("%d ",a[i]);
		}
	}
	return 0;
}