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 main(){
	int n;
	scanf("%d", &n);
	vector<int> a(n);
	for(int i = 0; i < n; i++)
		scanf("%d", &a[i]);
	vector<int> pre, nxt;
	pre = nxt = a;
	if(prev_permutation(pre.begin(), pre.end())) {
		for(int i = 0; i < n-1; i++)
			printf("%d ", pre[i]);
		printf("%d\n", pre[n-1]);
	}
	for(int i = 0; i < n-1; i++)
		printf("%d ", a[i]);
	printf("%d\n", a[n-1]);
	if(next_permutation(nxt.begin(), nxt.end())) {
		for(int i = 0; i < n-1; i++)
			printf("%d ", nxt[i]);
		printf("%d\n", nxt[n-1]);
	}
}