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;


int main(){
    int len_array;
    scanf("%d",&len_array);
    vector<int> int_array1(len_array);
    vector<int> int_array2(len_array);
    int temp;
    for(int i = 0;i < len_array;i++){
        scanf("%d",&temp),
        int_array1[i] = temp;
    }
    int_array2 = int_array1;
    if(prev_permutation(int_array1.begin(),int_array1.end())){
        for(int i = 0;i < int_array1.size();i++){
            if(i != int_array1.size()-1) printf("%d ",int_array1[i]);
            else printf("%d\n",int_array1[i]);
        }
    }
    for(int i = 0;i < int_array2.size();i++){
        if(i != int_array2.size()-1) printf("%d ",int_array2[i]);
        else printf("%d\n",int_array2[i]);
    }
    if(next_permutation(int_array2.begin(),int_array2.end())){
        for(int i = 0;i < int_array2.size();i++){
            if(i != int_array2.size()-1) printf("%d ",int_array2[i]);
            else printf("%d\n",int_array2[i]);
        }
    }


    return 0;   
}