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>
#define int long long
#define for0(i, n) for(int i = 0; i < (n); i++)
#define for1(i, n) for(int i = 1; i <= (n);i++)
#define all(x) x.begin(),x.end()
using namespace std;
vector<int>v;
int a[15], n;
int f() {
	int r = 0;
	for1(i, n) { r *= 10; r += a[i]; }
	return r;
}
void print(int p) {
	int x = 1;
	for0(i, n - 1)x *= 10;
	for0(i, n) {
		if (i)cout << ' ';
		cout << p / x; p %= x; x /= 10;
	}
	return;
}
signed main() {
	cin >> n;
	for1(i, n)a[i] = i;
	v.push_back(-1);
	do {
		v.push_back(f());
	} while (next_permutation(a + 1, a + n + 1));
	for1(i, n)cin >> a[i];
	int p = f();
	auto itr = lower_bound(all(v), p);
	itr--;
	if (*itr != -1) { print(*itr); cout << endl; }
	itr++;
	print(*itr); cout << endl;
	itr++;
	if (itr != v.end()) { print(*itr); cout << endl; }
}