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 <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <sstream>

#define INPUT_FROM_FILE (0) 
#if INPUT_FROM_FILE
#include <fstream>
#endif


int main(int argc, char **argv){
	int n;
	std::vector<int> a;
	std::vector<int> p;
	std::string buf;

#if INPUT_FROM_FILE
	std::ifstream ifs("test.txt");
	//FILE *fp = fopen("test.txt", "r");
#endif


#if INPUT_FROM_FILE
	ifs >> n;
#else
	std::cin >> n;
#endif

	a.resize(n);
	for(int i = 0; i < n; i++){
#if INPUT_FROM_FILE
		ifs >> a[i];
#else
		std::cin >> a[i];
#endif
	}

	p = a;
	if(std::prev_permutation(p.begin(), p.end())){
		std::cout << p[0];
		for(int i = 1; i < n; i++){
			std::cout << " " << p[i];
		}
		std::cout << std::endl;
	}
	std::cout << a[0];
	for(int i = 1; i < n; i++){
		std::cout << " " << a[i];
	}
	std::cout << std::endl;

	if(std::next_permutation(a.begin(), a.end())){
		std::cout << a[0];
		for(int i = 1; i < n; i++){
			std::cout << " " << a[i];
		}
		std::cout << std::endl;
	}

	return 0;
}