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;

typedef long long ll;
typedef long double ld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
typedef vector<ld> vld;
typedef vector<string> vstr;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef priority_queue<ll, vector<ll>, greater<ll>> spqll;  // 小さい順に取り出し
typedef priority_queue<ll, vector<ll>, less<ll>> bpqll;     // 大きい順に取り出し

#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define IREP(i, v) for (auto i = (v).begin(); i != (v).end(); ++i)
#define TS to_string
#define ALL(v) (v).begin(), (v).end()
#define endl "\n"

ll INF = 1e9;
ll MOD = 1000000007;
ll LINF = 1e18;
ld EPS = 1e-9;
ld PI = M_PI;
vll dx = {1, 0, -1, 0, 1, -1, -1, 1};
vll dy = {0, 1, 0, -1, 1, 1, -1, -1};
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b;}     //最小公倍数

void yes(){ cout << "Yes" << endl;}
void no(){ cout << "No" << endl;}

//-----------------------------------------

//-----------------------------------------
int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);

  ll n;
  cin >> n;
  vector<ll> vec(n);
  REP(i,n) vec[i] = i+1;
  vector<ll> mat(n);
  REP(i,n) cin >> mat[i];

  if(vec == mat){
    REP(i,n){
      if(i) cout << " ";
      cout << vec[i];
    }
    cout << endl;
    bool f = next_permutation(ALL(vec));
    if(!f) return 0;
    REP(i,n){
      if(i) cout << " ";
      cout << vec[i];
    }
    cout << endl;
    return 0;
  }


  while(1){
    vll tmp = vec;
    bool f = next_permutation(ALL(vec));
    if(!f) break;

    if(vec == mat){

       REP(i,n){
        if(i) cout << " ";
        cout << tmp[i];
      }
      cout << endl;

      REP(i,n){
        if(i) cout << " ";
        cout << vec[i];
      }
      cout << endl;

      bool g = next_permutation(ALL(vec));
      if(!g) break;

      REP(i,n){
        if(i) cout << " ";
        cout << vec[i];
      }
      cout << endl;

    }
  }

  return 0;
}