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 <iomanip>
#include<vector>
#include <algorithm>
#include <queue>
#include<string>
#include <map>
#include <cmath>
#include <deque>
#include <list>
#include <stack>
#include <queue>
#include <utility>
typedef long long ll;
typedef long double ld;


#define REP(i,s,n)for(ll i=s;i<(n);++i)
#define rep(i,n) REP(i,0,n)
using namespace std;
ll S[1000] = { 0 };
int N,M,Q;


int main() {
	cin >> N;
	vector<int>A(N);
	vector<int>B(N);
	vector<int>C(N);
		rep(i, N) {
			cin >> A[i];
			B[i] = A[i];
			C[i] = A[i];
		}
		if (prev_permutation(A.begin(), A.end())) {
			rep(i, N - 1) {
				cout << A[i] << " ";
			}
			cout << A[N - 1] << endl;
		}
			rep(i, N - 1) {
				cout << B[i] << " ";
			}
			cout << B[N - 1] << endl;
		if (next_permutation(C.begin(), C.end())) {
			rep(i, N - 1) {
				cout << C[i] << " ";
			}
			cout << C[N - 1] << endl;
		}
	return 0;
}