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>
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define Rep(i,a,b) for(int i=a;i<b;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define rev(i,n) for(int i=n-1;i>=0;i--)
#define vi vector<int>
#define pi pair<int,int>
#define vp vector<pair<int,int> >
#define all(v) v.begin(),v.end()
using namespace std;

signed main(){
    int n;
    cin>>n;
    vi a(n);
    rep(i,n)cin>>a[i];
    vi b(n),c(n);
    copy(all(a),b.begin());
    copy(all(a),c.begin());
    if(prev_permutation(all(b))){
        rep(i,n)cout<<(i?" ":"")<<b[i];
        cout<<endl;
    }
    rep(i,n)cout<<(i?" ":"")<<a[i];
    cout<<endl;
    if(next_permutation(all(c))){
        rep(i,n)cout<<(i?" ":"")<<c[i];
        cout<<endl;
    }
    return 0;
}