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;

void output(vector<int> v) {
	for (int i = 0; i < v.size(); i++) {
		if (i) printf(" ");
		printf("%d", v[i]);
	}
	printf("\n");
}

int main(void) {
	int n, tmp;
	vector<int> v;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &tmp);
		v.push_back(tmp);
	}
	bool prev_flag;
	if (prev_permutation(v.begin(), v.end())) {
		output(v);
	}
	next_permutation(v.begin(), v.end());
	output(v);
	if (next_permutation(v.begin(), v.end())) {
		output(v);
	}
}