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
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

set<int> s;
map<int, int> mp;

int main() {
  int N; cin >> N;
  vector<int> a(N); 
  rep(i, N) {
    cin >> a[i];
    a[i]--;
  }
  vector<int> ans;

  rep(i, N) s.insert(i);
  rep(i, N) mp[a[i]]++;

  int ng = -1;
  rep(i, N) {
    int top = -1;
    for (int x : s) {
      if (x == ng) continue;
      int tmp = a[x];
      if (s.count(tmp) == 1 and mp[tmp] == N - i - 1) continue;
      if (s.size() == 3) {
        vector<int> test;
        for (auto y : s) {
          if (x == y) continue;
          test.push_back(y);
        }
        if (a[test[0]] == test[1] and a[test[1]] == test[0]) continue;
      }
      top = x;
      break;
    }
    if (top == -1) {
      cout << -1 << endl;
      return 0;
    }
    ans.push_back(top);
    s.erase(top);
    mp[a[top]]--;
    ng = a[top];
  }

  rep(i, N) {
    cout << ans[i] + 1 << " ";
  }
  cout << endl;
}