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;
#define rep(i,n) for(int i=0;i<(n);i++)

typedef long long ll;
typedef pair<int, int> P;

#define MAX 100005
#define INF 1001001001

void out(vector<int> a) {
  int n = a.size();
  rep(i,n) {
    cout << a[i];
    if (i==n-1) cout << endl;
    else cout << " ";
  }
}
int main(int, char**)
{
  int n;
  cin >> n;
  vector<int> a(n);
  rep(i,n) cin >> a[i];
  if(prev_permutation(a.begin(), a.end())) out(a);
  next_permutation(a.begin(), a.end()); out(a);
  if(next_permutation(a.begin(), a.end())) out(a);

  return 0;
}