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<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>
#include<bitset>
#include<limits.h>
#include<cassert>
#include<iterator>
#include<complex>
#include<stack>
#include<sstream>
#include<iterator>
#include<list>
using namespace std;
typedef  long long int lint;
typedef pair<int, int> IP;
typedef pair<lint, lint> LLP;
typedef pair<char, char>CP;
#define rep(i, n) for (int i = 0; i < n; i++)
#define repr(i, n) for (int i = n; i >= 0; i--)
#define sort(v) sort((v).begin(), (v).end())
#define reverse(v) reverse((v).begin(), (v).end())
#define upper(v,hoge) upper_bound(v.begin(),v.end(),hoge)
#define lower(v,hoge) lower_bound(v.begin(),v.end(),hoge)
#define llower(v,hoge) *lower_bound(v.begin(), v.end(), hoge)
#define lupper(v,hoge) *upper_bound(v.begin(), v.end(), hoge)

int main() {
	int N;
	cin >> N;
	vector<int>A(N);
	rep(i, N) {
		cin >> A[i];
	}
	vector<int>AA(N);
	rep(i, N) {
		AA[i] = A[i];
	}
	sort(A);
	vector<vector<int>>B;
	do {
		B.push_back(A);
	} while (next_permutation(A.begin(), A.end()));
	int it = 0;
	rep(i, B.size()) {
		if (B[i] == AA) {
			it = i;
		}
	}
	if (0 <= it - 1) {
		rep(i, N) {
			cout << B[it - 1][i];
			if (i != N - 1) {
				cout << ' ';
			}
			else {
				cout << endl;
			}
		}
	}
	rep(i, N) {
		cout << B[it][i];
		if (i != N - 1) {
			cout << ' ';
		}
		else {
			cout << endl;
		}
	}
	if (it + 1<B.size()) {
		rep(i, N) {
			cout << B[it + 1][i];
			if (i != N - 1) {
				cout << ' ';
			}
			else {
				cout << endl;
			}
		}
	}
}