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:


/**
 *      purpose : 
 *      author  : kyomukyomupurin
 *      created : 
**/

// input/output
#include <iostream>
#include <fstream>
#include <sstream>
// container class
#include <array>
#include <deque>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <vector>
// math, algorithm
#include <algorithm>
#include <cmath>
#include <complex>
#include <numeric>
// etc
#include <bitset>
#include <cassert>
#include <cctype>
#include <cstring>
#include <chrono>
#include <iomanip>
#include <random>
#include <utility>
// using-directive
using namespace std;
// alias template
using int64 = long long;
using vi = vector<int>;
using vl = vector<int64>;
using pii = pair<int, int>;
using pll = pair<int64, int64>;
// text macro replacement
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define print(x) cout << (x) << '\n'
#define debug(x) cerr << #x << ": " << (x) << '\n'
#define dbg(v) for (size_t _ = 0; _ < v.size(); ++_){cerr << #v << "[" << _ << "] : " << v[_] << '\n';}
// variadic template
template<typename T> inline void chmin(T &a, T b) {if (a > b) a = b; return;}
template<typename T> inline void chmax(T &a, T b) {if (a < b) a = b; return;}
// constant
const int INF = (1<<30) - 1;
const int64 INF64 = (1LL<<62) - 1;
const int MOD = 1000000007;

int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);

  int n; cin >> n;
  vi a(n); rep(i, n) cin >> a[i];

  vi b = a;
  vi c = a;

  if (prev_permutation(all(b))) {
    for (int i = 0; i < n; ++i) {
      cout << b[i];
      if (i != n - 1) cout << " ";
    }
    cout << '\n';
  }

  for (int i = 0; i < n; ++i) {
    cout << a[i];
    if (i != n - 1) cout << " ";
  }
  cout << '\n';

  if (next_permutation(all(c))) {
    for (int i = 0; i < n; ++i) {
      cout << c[i];
      if (i != n - 1) cout << " ";
    }
    cout << '\n';
  }


  return 0;
}