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
//------------------------------------------
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <fstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
#include <ctime>


using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }

//math
//-------------------------------------------
template<class T> inline T sqr(T x) { return x * x; }

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define EXISTch(s,c) ((((s).find_first_of(c)) != std::string::npos)? 1 : 0)//cがあれば1 if(1)
#define SORT(c) sort((c).begin(),(c).end())

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  FOR(i,0,n)
#define loop(n) FOR(i,0,n)
#define rrep(i,a,b) for(int i=(a);i>=(b);--i)
//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI = acos(-1.0);
const int INF = (int)1000000007;
const LL MOD = (LL)1000000007;//10^9+7
const LL INF2 = (LL)100000000000000000;//10^18

int main() {
	int n; cin >> n;

	vector<int> a(n);

	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}

	vector<int> b = a;

	if (prev_permutation(a.begin(), a.end())) {
		for (int i = 0; i < n; i++) {
			cout << a[i];

			if (i == n - 1) {
				cout << endl;
			}
			else {
				cout << " ";
			}
		}

		
	}

	

	
	for (int i = 0; i < n; i++) {
		cout << b[i];

		if (i == n - 1) {
			cout << endl;
		}
		else {
			cout << " ";
		}
	}
	
	
	
	if (next_permutation(b.begin(), b.end())) {
		for (int i = 0; i < n; i++) {
			cout << b[i];

			if (i == n - 1) {
				cout << endl;
			}
			else {
				cout << " ";
			}
		}
	}
	return 0;
}