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()
{
   vector<int>a;
    int i,j,b,c,n,p,q,r,t,x,k;

    cin>>n;


for(i=0;i<n;i++)
{
    cin>>x;
    a.push_back(x);
}
if(!is_sorted(a.begin(),a.end()))
   {
       prev_permutation(a.begin(),a.end());

for(i=0;i<n;i++)
{
    if(i!=n-1)
    {
         cout<<a[i]<<" ";
    }
 else
 {
     cout<<a[i];
 }
}
    cout<<endl;
}
else
{
    prev_permutation(a.begin(),a.end());
}

next_permutation(a.begin(),a.end());
for(i=0;i<n;i++)
{
       if(i!=n-1)
    {
         cout<<a[i]<<" ";
    }
 else
 {
     cout<<a[i];
 }
}
cout<<endl;
if(!is_sorted(a.begin(),a.end(),greater<int>()))
   {
       next_permutation(a.begin(),a.end());

for(i=0;i<n;i++)
{
     if(i!=n-1)
    {
         cout<<a[i]<<" ";
    }
 else
 {
     cout<<a[i];
 }
}
    cout<<endl;
   }


return 0;

}