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 range(i, m, n) for(int i = m; i < n; i++)
#define husk(i, m, n) for(int i = m; i > n; i--)

int n;

int main() {
  cin >> n;
  vector<int> cnt(n);
  vector<int> a(n);
  range(i, 0, n) {
    cin >> a[i];
    a[i]--;
    cnt[a[i]]++;
  }
  vector<int> pos;
  range(i, 0, n) if(cnt[i] == n - 1) {
    pos.push_back(i);
  }
  if(pos.size() > 1) {
    cout << -1;
    return 0;
  }
  if(!pos.empty()) {
    vector<int> res = {pos[0]};
    range(i, 0, n) if(i != pos[0]) res.push_back(i);
    if(a[res[0]] == res[1]) swap(res[1], res[2]);
    for(int i : res) cout << i + 1 << " ";
    return 0;
  }
  set<int> cur;
  vector<int> res;
  range(i, 0, n) cur.insert(i);
  while(cur.size() > 1) {
    auto it = cur.begin();
    if(!res.empty() && a[res.back()] == *it) it = next(it);
    res.push_back(*it);
    cur.erase(it);
  }
  int c = *cur.begin();
  cur.erase(c);
  while(res.size()) {
    if(a[res.back()] == c) {
      cur.insert(res.back());
      res.pop_back();
    }
    else break;
  }
  if(cur.size() == 1 && a[c] == *cur.begin()) {
    res.push_back(*cur.begin());
    res.push_back(c);
    auto check = [&]() {
      range(i, 0, n - 1) if(a[res[i]] == res[i + 1]) return false;
      return true;
    };
    sort(res.begin() + n - 3, res.end());
    while(!check()) {
      assert(next_permutation(res.begin() + n - 3, res.end()));
    }
  } else {
    res.push_back(c);
    while(cur.size()) {
      int c = *cur.begin();
      if(a[res.back()] == c) c = *next(cur.begin());
      cur.erase(c);
      res.push_back(c);
    }
  }
  for(int i : res) cout << i + 1 << " ";
  return 0;
}
