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;
typedef long long ll;

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> indeg(n);
  for (int&ai : a) {
    cin >> ai;
    ai--;
    indeg[ai]++;
  }

  set<int> left;
  set<pair<int,int>> deg;
  for (int i = 0; i < n; i++) {
    left.insert(i);
    deg.insert({indeg[i], i});
  }
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    if (deg.rbegin()->first == n-1-i) {
      ans.push_back(deg.rbegin()->second);
      deg.erase(*deg.rbegin());
      if (deg.size() && deg.rbegin()->first == n-1-i) {
	assert(n == 2);
	cout << -1 << endl;
	return 0;
      }
    } else {
      auto it = left.begin();
      if (ans.size() && *it == a[ans.back()]) ++it;
      ans.push_back(*it);
    }
    {
      int p = ans.back();
      //cout << p+1 << endl;
      left.erase(p);
      deg.erase({indeg[p],p});
      if (left.count(a[p])) {
	deg.erase({indeg[a[p]],a[p]});
	deg.insert({--indeg[a[p]],a[p]});
      }

      if (left.size() == 2) {
	int x = *left.begin(), y = *left.rbegin();
	if (a[x] == y && a[y] == x) {
	  int z = ans.back();
	  ans.pop_back();
	  if (!(ans.size() && a[ans.back()] == x || a[z] == y)) {
	    ans.push_back(x);
	    ans.push_back(z);
	    ans.push_back(y);
	  } else {
	    ans.push_back(y);
	    ans.push_back(z);
	    ans.push_back(x);
	  }
	  break;
	}
      }
    }
  }
  for (int i : ans) cout << i+1 << ' ';
  cout << endl;
}
