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 <iostream>
#include <cstring>
using namespace std;

char S[10], A[10], D[3][10];
bool F[10];

bool dfs(int pos, bool &flag, const int N) {
  if (pos == N) {
    if (flag) {
      strncpy(D[2], S, N);
      return true;
    }
    if (!strcmp(S, A)) flag = true;
    strncpy(D[0], D[1], N);
    strncpy(D[1], S, N);
  } else {
    for (int i = 0; i < N; ++i) {
      if (F[i]) continue;
      S[pos] = (i + 1) + '0';
      F[i] = true;
      bool ret = dfs(pos + 1, flag, N);
      if (ret) return true;
      F[i] = false;
    }
  }
  return false;
}

int main() {
  int N;
  cin >> N;

  for (int i = 0; i < N; ++i) {
    cin >> A[i];
  }

  bool flag = false;
  dfs(0, flag, N);

  for (int i = 0; i < 3; ++i) {
    if (!strlen(D[i])) continue;
    for (int j = 0; j < N; ++j) {
      if (j > 0) cout << ' ';
      cout << D[i][j];
    }
    cout << endl;
  }

  return 0;
}