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;
#define rep(i,n) for(int i=0;i<n;i++)
int n;
vector<vector<int>>v;
vector<int>a, c;
int kaijo(int x) {
	int ans = 1;
	for (int i = 1; i <= x; i++)ans *= i;
	return ans;
}
int main() {
	cin >> n;
	a.resize(n);
	c.resize(n);
	rep(i, n)cin >> a[i];
	if (n == 1) {
		puts("1");
		return 0;
	}
	rep(i, n)c[i] = i + 1;
	do {
		v.push_back(c);
	} while (next_permutation(c.begin(), c.end()));
	rep(i, v.size()) {
		if (v[i] == a) {
			if (i == 0) {
				rep(j, v[i].size())cout << v[i][j] << (j == v[i].size() - 1 ? "\n" : " ");
				i++;
				rep(j, v[i].size())cout << v[i][j] << (j == v[i].size() - 1 ? "\n" : " ");
				return 0;
			}
			else if (i == v.size() - 1) {
				i--;
				rep(j, v[i].size())cout << v[i][j] << (j == v[i].size() - 1 ? "\n" : " ");
				i++;
				rep(j, v[i].size())cout << v[i][j] << (j == v[i].size() - 1 ? "\n" : " ");
				return 0;
			}
			else {
				i--;
				rep(j, v[i].size())cout << v[i][j] << (j == v[i].size() - 1 ? "\n" : " ");
				i++;
				rep(j, v[i].size())cout << v[i][j] << (j == v[i].size() - 1 ? "\n" : " ");
				i++;
				rep(j, v[i].size())cout << v[i][j] << (j == v[i].size() - 1 ? "\n" : " ");
				return 0;
			}
		}
	}
	return 0;
}