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,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=s;i<n;i++)
#define all(a) a.begin(),a.end()
typedef long long ll;
const ll inf = 1e9;
int main(){
    int n; cin >> n;
    vector<ll> a(n), b(n);
    rep(i,n)cin >> a[i];
    b = a;
    if(prev_permutation(all(b)))
        rep(i,n)cout << b[i] << ((i < n-1)?" ":"\n");
    rep(i,n)cout << a[i] << ((i < n-1)?" ":"\n");
    if(next_permutation(all(a)))
        rep(i,n)cout << a[i] << ((i < n-1)?" ":"\n");
    
    return 0;
}