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,j,k) for(int i = (int)j;i <= (int)k;i ++)
#define debug(x) cerr<<#x<<":"<<x<<endl
const int maxn=(int)1e6+5;
int a[10]; int n;
void print(){
    rep(i,0,n-1)cout<<a[i]<<(i==n-1?'\n':' ');
}
    
int main(){
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
   cin>>n;
    rep(i,0,n-1){
        cin>>a[i];
    }
    if(prev_permutation(a,a+n))print();
    next_permutation(a,a+n);
    print();
    if(next_permutation(a,a+n))print();
}
/* 
3
1 2 3
3
1 3 2
 */