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;
//#include <boost/multiprecision/cpp_int.hpp>
//using multiInt = boost::multiprecision::cpp_int;

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename Q_type>
using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>;

const int MOD_TYPE = 1;
const ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
const int INF = (int)1e9;
const ll LINF = (ll)4e18;
const double PI = acos(-1.0);

#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << endl
#define possible(n) cout << ((n) ? "possible" : "impossible") << endl
#define Yay(n) cout << ((n) ? "Yay!" : ":(") << endl
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << endl;

vector<int> Dx = {0, 0, -1, 1, -1, 1, -1, 1, 0};
vector<int> Dy = {1, -1, 0, 0, -1, -1, 1, 1, 0};

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(50) << setiosflags(ios::fixed);
  int n;
  cin >> n;
  vector<int> a(n), b(n), c(n);
  rep(i, n)
  {
    cin >> a[i];
    b[i] = c[i] = a[i];
  }
  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 (NP(c))
  {
    rep(i, n) cout << c[i] << (i == n - 1 ? "\n" : " ");
  }
  return 0;
}