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:
/*input
3
2 1 3
*/
#include <iostream>
#include <algorithm>

int main(int argc, char const *argv[]){
	int n;
	std::cin>>n;
	int arr[n];
	for(int i = 0; i < n; i++){
		std::cin>>arr[i];
	}
	int arr1[n], arr2[n];
	std::copy(arr, arr+n, arr1);
	std::copy(arr, arr+n, arr2);
	if(std::prev_permutation(arr1, arr1+n)){
		for(int i = 0; i < n; i++){
			std::cout<<arr1[i];
			if(i != n-1){
				std::cout<<' ';	
			}
		}
		std::cout<<std::endl;
	}
	for(int i = 0; i < n; i++){
		std::cout<<arr[i];
		if(i != n-1){
			std::cout<<' ';	
		}
	}
	std::cout<<std::endl;
	if(std::next_permutation(arr2, arr2+n)){
		for(int i = 0; i < n; i++){
			std::cout<<arr2[i];
			if(i != n-1){
				std::cout<<' ';	
			}
		}
		std::cout<<std::endl;
	}
	return 0;
}