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 <vector>
#include <algorithm>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <numeric>
#include <climits>
#include <iterator>
#include <iomanip>
#include <stack>
#include <set>
#include <cstdio>
#include <bitset>
#include <functional>
#include <tuple>
#include <list>
#include <deque>
using namespace std;
typedef pair<int, int> P;

#define FOR(i, a, n) for (ll i = (ll)a; i<(ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)

typedef long long ll;
typedef vector<int> VI;


struct More {
	bool operator()(const pair<int, int>& x, const pair<int, int>& y) const {
		return x.first > y.first;
	}
};


int a[10];
vector<int> v;
int main() {
    cin.tie(0);
   	ios::sync_with_stdio(false);
    
    int n; cin >> n;
    char s;
    string t="";
    REP(i, n){
        cin >> s;
        t += s;
        v.push_back((int)s-'0');
    }
    sort(v.begin(), v.end());
    
    
    vector<string> ans;
    do{
        string str="";
        for(int i=0; i<n; ++i){
            str += (char)v[i] +'0';
        }
        ans.push_back(str);
    }while(next_permutation(v.begin(), v.end()));
           
    for(int i=0; i<ans.size(); ++i){
        if(ans[i]==t){
            if(i>0) {
                for(int j=0; j<ans[i-1].size(); ++j){
                    if(j) cout << " ";
                    cout << ans[i-1][j];
                }
                cout << endl;
            }
            for(int j=0; j<ans[i].size(); ++j){
                    if(j) cout << " ";
                    cout << ans[i][j];
               
            } cout << endl;
            if(i<ans.size()-1){
                
                for(int j=0; j<ans[i+1].size(); ++j){
                    if(j) cout << " ";
                    cout << ans[i+1][j];
                }
                cout << endl;
            }
            break;
        }
    }
	return 0;
}