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 <iostream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;

void print(const vector<int>& res) {
  for (int i = 0; i < res.size(); i++)
    cout << (i == 0 ? "" : " ") << res[i];
  cout << endl;
}

bool check(const vector<int>& res, const vector<int>& G){
  for (int i = 0; i < res.size()-1; i++)
    if (res[i+1] == G[res[i]]) return false;
  return true;
}

int main(){
  int N; cin >> N;
  vector<int> G(N+1), in(N+1, 0);
  set<int> av;
  for (int i = 1; i <= N; i++) {
    cin >> G[i];
    av.insert(i);
    in[G[i]]++;
  }

  if (N == 2) {
    cout << -1 << endl;
    exit(0);
  }

  int u = -1;
  vector<int> res;
  while (not av.empty()) {
    auto it = av.begin();
    if (u != -1 and *it == G[u]) ++it;
    if (it == av.end()) --it;
    u = *it;

    auto it2 = av.find(G[u]);
    if (it2 != av.end() and in[G[u]] == av.size()-1) u = G[u];

    res.push_back(u);
    av.erase(u);
    in[G[u]]--;
  }

  int& a = res[N-1];
  int& b = res[N-2];
  int& c = res[N-3];
  int sum = a+b+c;
  c = min(a, min(b, c));
  a = max(a, max(b, c));
  b = sum - c - a;

  do {
    if (check(res, G)) {
      print(res);
      exit(0);
    }
  } while (next_permutation(res.begin(), res.end()));
}
