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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;

#define r(i,n) for(int i=0;i<n;i++)

int main()
{
	int n, i;
	scanf("%d", &n);
	vector<int> a(n),b(n);
	//vector<int> cpp[11];
	r(i, n) scanf("%d", &a[i]),b[i]=a[i];

	if (prev_permutation(a.begin(),a.end()))r(i,n) printf(i < n - 1 ? "%d " : "%d\n", a[i]);
	r(i,n) printf(i < n - 1 ? "%d " : "%d\n", b[i]);
	if (next_permutation(b.begin(), b.end())) r(i,n) printf(i < n - 1 ? "%d " : "%d\n", b[i]);

	return 0;


}