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;

void print(vector<int> vec){
  for(int i=0; i<vec.size(); i++){
    if(i!=0) cout <<" ";
    cout << vec[i];
  }
  cout << endl;
}

int main() {
  int n; cin >> n;
  vector<int> a(n);
  for(int i=0; i<n; i++) cin >> a[i];

  int N=1;
  for(int i=1; i<=n; i++) N*=i;

  vector<int> v(n);
  iota(v.begin(),v.end(),1);
  vector<vector<int> > b(400000);
  int i=0, k=0;
  do {
    for(auto x:v){
      b[i].push_back(x);
      //cout << x <<" ";
    }
    //cout << endl;
    bool f=true;
    for(int j=0; j<n; j++){
      if(b[i][j]!=a[j]) f=false;
    }
    if(f){
      k=i;
    }
    i++;
  } while(next_permutation(v.begin(),v.end()));

  //cout << k <<" "<< N << endl;
  if(N==1) print(b[k]);
  else if(k==0){
    print(b[k]);
    print(b[k+1]);
  }
  else if(k==N-1){
    print(b[k-1]);
    print(b[k]);
  }
  else {
    print(b[k-1]);
    print(b[k]);
    print(b[k+1]);
  }

}