Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

4
2 3 4 1

Sample Output 1

1 3 2 4

The arrangement (1,2,3,4) is lexicographically smaller than (1,3,2,4), but is invalid, since it violates the condition "to the immediate right of Card 1 is not Card 2."

Sample Input 2

2
2 1

Sample Output 2

-1

If no arrangements satisfy the conditions, print -1.

Sample Input 3

13
2 3 4 5 6 7 8 9 10 11 12 13 12

Sample Output 3

1 3 2 4 6 5 7 9 8 10 12 11 13

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
template<typename T>
void fin(T a){
  cout<<a<<endl;
  exit(0);
}
const int INF=1e9;

signed main(){
  int n;cin>>n;
  vector<int> v(n);
  for(int i=0;i<n;i++)cin>>v[i];
  if(n==2)fin(-1);
  map<int,int> m;
  for(int i=0;i<n;i++)m[v[i]]++;
  vector<int> ans;
  set<int> s;
  for(int i=1;i<=n;i++)s.insert(i);
  while(ans.size()<n){
    if(ans.size()+11>=n){
      int kaka=n-ans.size();
      vector<int> w(kaka);
      for(int i=0;i<kaka;i++){
        w[i]=*s.begin();s.erase(w[i]);
      }
      //cout<<"?";for(int i=0;i<3;i++)cout<<w[i]<<" ";cout<<endl;
      do{
        //cout<<"??";for(int i=0;i<3;i++)cout<<w[i]<<" ";cout<<endl;
        bool f=true;
        int T=-1;
        if(ans.size()>0)T=v[ans[ans.size()-1]-1];
        for(int i=0;i<kaka;i++){
          if(i==0){
            if(T==w[0])f=false;
          }
          else{
            if(v[w[i-1]-1]==w[i])f=false;
          }
        }
        if(f){
          for(int i=0;i<kaka;i++)ans.push_back(w[i]);
          break;
        }
      }while(next_permutation(w.begin(),w.end()));
      break;
    }
    int p=*s.begin();
    /*if(ans.size()==n-1){
      if(p==v[ans[ans.size()-1]-1])fin(-1);
      ans.push_back(p);break;
    }*/
    int t=v[p-1];
    int ho=-1;
    if(ans.size())if(v[ans[ans.size()-1]-1]==p){
      ho=p;s.erase(p);s.insert(INF);
      p=*s.begin();if(p==INF)fin(-1);
      t=v[p-1];
    }
    if(s.count(t)&&m[t]+1==s.size()){
      s.erase(p);
      int q=*s.begin();
      ans.push_back(q);
      s.erase(q);
      if(--m[v[q-1]]==0)m.erase(v[q-1]);
      s.insert(p);
    }
    else{
      ans.push_back(p);
      s.erase(p);
      if(--m[t]==0)m.erase(t);
    }
    if(~ho){
      s.erase(INF);s.insert(ho);
    }
  }
  if(ans.size()<n)fin(-1);
  for(int p:ans)cout<<p<<" ";cout<<endl;
}