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
//Link : https://atcoder.jp/contests/dwacon6th-prelims/tasks/dwacon6th_prelims_d

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define N 100005
int a[N];
int indeg[N];
void solve() {
  int n;scanf("%d ", &n);
  for(int i=1;i<=n;++i) {
    scanf("%d ", &a[i]);
    indeg[a[i]]++;
  }
  set<pair<int,int> >s;
  set<int> remNode;
  for(int i=n;i;--i) {
    s.insert(make_pair(indeg[i],i));
    remNode.insert(i);
  }
  if(n==2) {
    printf("-1\n");
    return;
  }
  int ban = -1;
  while(true) {
    if(n<=3) {
      vector<int> v;
      for(int node : remNode) {
        v.push_back(node);
      }
      vector<int> ret;
      do {
        if(v[0]==ban) {
          continue;
        }
        bool isGood = true;
        for(int i=0;i<v.size()-1;++i) {
          if(a[v[i]]==v[i+1]) {
            isGood = false;
            break;
          }
        }
        if(isGood) {
          if(ret.size()==0) {
            ret = v;
          } else {
            ret = min(ret,v);
          }
        }
      } while(next_permutation(v.begin(),v.end()));
      for(int i=0;i<ret.size();++i) {
        printf("%d ", ret[i]);
      }
      putchar('\n');
      break;
    } else {
      pair<int,int> tmp = *prev(s.end());
      int chNode = -1;
      if(tmp.first==n-1) {
        chNode = tmp.second;
      } else {
        for(int node : remNode) {
          if(node!=ban) {
            chNode = node;
            break;
          }
        }
      }
      printf("%d ", chNode);
      ban = a[chNode];
      s.erase(make_pair(indeg[chNode],chNode));
      remNode.erase(chNode);
      if(remNode.count(a[chNode])) {
        s.erase(make_pair(indeg[a[chNode]],a[chNode]));
        --indeg[a[chNode]];
        s.insert(make_pair(indeg[a[chNode]],a[chNode]));
      }
      --n;
    }
  }
}

int main() {
    //freopen("input.txt","r",stdin);
    solve();
    return 0;
}
