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 <bits/stdc++.h>

using namespace std;
typedef long long ll;

// #define DEBUG

vector<string> perm_enum(string s, int n)
{
    vector<string> a;
    string t;
    if (n == 1)
    {
        a.push_back(s);
        return a;
    }
    else
    {
        for (int i = 0; i < n; i++)
        {
            t = "";
            for (int j = 0; j < n; j++)
            {
                if (i != j)
                    t += s[j];
            }
            vector<string> b = perm_enum(t, n - 1);
            for (string r : b)
                a.push_back(s[i] + r);
        }

        return a;
    }
}

void printperm(string s, int n)
{
    for (int i = 0; i < n - 1; i++)
        cout << s[i] << " ";
    cout << s[n - 1] << endl;
}

int frac(int n){
    int ret = 1;
    for (int i = 1; i <= n; i++)
        ret *= i;
    return ret;
}

int main()
{

#ifdef DEBUG
    cout << "DEBUG MODE" << endl;
    ifstream in("input.txt"); //for debug
    cin.rdbuf(in.rdbuf());    //for debug
#endif
    int n, index;
    cin >> n;
    string t, target;
    char x;
    for (int i = 0; i < n; i++){
        cin >> x;
        target += x;
    }

    for (int i = 0; i < n; i++)
        t += i + '1';
    vector<string> ans = perm_enum(t, n);
    for (int i = 0; i < frac(n); i++)
        if (ans[i] == target)
            index = i;

    for (int i = -1; i <= 1; i++)
        if (0 <= index+i && index+i <frac(n))
            printperm(ans[index+i],n);
    
    return 0;
}