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 rep(i ,n) for(int i=0;i<(int)(n);++i)
using namespace std;
typedef long long int int64;
typedef unsigned long long uint64;

int main(){
    int n; cin >> n;
    vector<int> a(n);
    vector<int> m(n);
    rep(i, n) cin >> a[i];
    rep(i, n) m[i]=a[i];
    if(!prev_permutation(a.begin(),a.end())) {}
    else {
        rep(i ,n - 1) cout << a[i] << " ";
        cout << a[n-1] << endl;
    }
    rep(i ,n - 1) cout << m[i] << " ";
    cout << m[n-1] << endl;
    if(!next_permutation(m.begin(),m.end())) {}
    else {
        rep(i ,n - 1) cout << m[i] << " ";
        cout << m[n-1] << endl;
    }
}