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 n;cin>>n;
   vector<int > v(n);
    for (int i = 0; i < n; ++i)
        cin>>v[i];
    vector <int > p(v);
    if(prev_permutation(p.begin(),p.end())){
        for (int i = 0; i < n-1; ++i) {
            cout<<p[i]<<" ";
        }
        cout<<p[n-1]<<endl;
    }
        for (int i = 0; i < n-1; ++i) {
            cout<<v[i]<<" ";
        }
        cout<<v[n-1]<<endl;
    if(next_permutation(v.begin(),v.end())){
        for (int i = 0; i < n-1; ++i) {
            cout<<v[i]<<" ";
        }
        cout<<v[n-1]<<endl;
    }
    return 0;
}