Niwango has N cards, numbered 1,2,\ldots,N. He will now arrange these cards in a row.

Niwango wants to know if there is a way to arrange the cards while satisfying all the N conditions below. To help him, determine whether such a way exists. If the answer is yes, also find the lexicographically smallest such arrangement.

* To the immediate right of Card 1 (if any) is NOT Card a_1.
* To the immediate right of Card 2 (if any) is NOT Card a_2.
* \vdots
* To the immediate right of Card N (if any) is NOT Card a_N.

Constraints

* 2 \leq N \leq 10^{5}
* 1 \leq a_i \leq N
* a_i \neq i

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 \ldots a_N


Output

If no arrangements satisfy the conditions, print `-1`. If such arrangements exist, print the lexicographically smallest such arrangement, in the following format:


b_1 b_2 \ldots b_N


Here, b_i represents the i-th card from the left.

Examples

Input

4
2 3 4 1


Output

1 3 2 4


Input

2
2 1


Output

-1


Input

13
2 3 4 5 6 7 8 9 10 11 12 13 12


Output

1 3 2 4 6 5 7 9 8 10 12 11 13
/**
 *    author:  tourist
 *    created: 11.01.2020 14:33:51       
**/
#include <bits/stdc++.h>
 
using namespace std;
 
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> deg(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    --a[i];
    deg[a[i]] += 1;
  }
  vector<int> b(n, -1);
  vector<int> used(n, 0);
  function<void(int, int)> Dfs = [&](int v, int mn) {
    if (v == n) {
      for (int i = 0; i < n; i++) {
        if (i > 0) {
          cout << " ";
        }
        cout << b[i] + 1;
      }
      cout << '\n';
      exit(0);
    }
    if (v > 0) {
      int who = a[b[v - 1]];
      if (!used[who] && deg[who] == n - v - 1) {
        return;
      }
    }
    while (used[mn]) {
      ++mn;
    }
    for (int i = mn; i < n; i++) {
      if (used[i]) {
        continue;
      }
      if (v > 0 && i == a[b[v - 1]]) {
        continue;
      }
      int new_mn = mn;
      if (i == mn) {
        ++new_mn;
      }
      b[v] = i;
      --deg[a[i]];
      used[i] = 1;
      Dfs(v + 1, new_mn);
      used[i] = 0;
      ++deg[a[i]];
      b[v] = -1;
    }
  };
  Dfs(0, 0);
  cout << -1 << '\n';
  return 0;
}
