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;
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) rep(i,0,n)

int main()
{
    int n;
    cin>>n;
    vector<int>a(n);
    REP(i,n)cin>>a[i];
    if(prev_permutation(a.begin(),a.end())){REP(i,n){if(i!=0)cout<<" ";cout<<a[i];}cout<<endl;}
    next_permutation(a.begin(),a.end());
    REP(i,n){if(i!=0)cout<<" ";cout<<a[i];}cout<<endl;
    if(next_permutation(a.begin(),a.end())){REP(i,n){if(i!=0)cout<<" ";cout<<a[i];}cout<<endl;}
}