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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<n;i++)

/* 大文字を小文字に変換 */
char tolower(char c) {return (c + 0x20);}
/* 小文字を大文字に変換 */
char toupr(char c) {return (c - 0x20);}

// if('A'<=s[i] && s[i]<='Z') s[i] += 'a'-'A';

/*
string s = "abcdefg"
s.substr(4) "efg"
s.substr(0,3) "abc"
s.substr(2,4) "cdef"
*/

int main()
{
    int n; cin >> n;
    vector<int> A;
    rep(i,n){
        int Ai; cin >> Ai;
        A.push_back(Ai);
    }
    auto B = A;
    sort(A.begin(), A.end());
    vector<int> ori_A;
    rep(i,n) ori_A.push_back(-1);

    bool is = false;
    bool is2 = false;
    do{
        if(A == B){
            if(ori_A[0] != -1){
                rep(i,n){
                    cout << ori_A[i];
                    if(i != n-1) cout << " ";
                }
                cout << endl;
            }
            rep(i,n){
                cout << A[i];
                if(i != n-1) cout << " ";
            }
            cout << endl;
            is = true;
        }
        ori_A = A;
        if(is2){
            rep(i,n){
                cout << A[i];
                if(i != n-1) cout << " ";
            }
            cout << endl;
            break;
        }
        if(is) is2 = true;
    }while(next_permutation(A.begin(), A.end()));
}