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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#define long long long
#define LF '\n'
#define ALL(x) x.begin(), x.end()
using namespace std;
typedef pair<int,int> pii;
template<class A, class B>inline bool chmax(A &a, const B &b){return b>a ? a=b,1 : 0;}
template<class A, class B>inline bool chmin(A &a, const B &b){return b<a ? a=b,1 : 0;}
constexpr int INF = 0x3f3f3f3f;

template<class T>
void dump(const T &a) {
  for (int i = 0 ; i < a.size(); ++i) {
    cout << a[i] << (i+1 < a.size() ? ' ' : LF);
  }
}

signed main()
{
  cin.tie(nullptr), ios::sync_with_stdio(false);

  int N; cin >> N;
  vector<int> a(N);

  for (int i = 0; i < N; ++i) cin >> a[i];

  vector<int> b = a;

  if (prev_permutation(ALL(a))) dump(a);
  dump(b);
  if (next_permutation(ALL(b))) dump(b);

  return 0;
}