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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <iostream>
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
using namespace std;
using ll = long long int;
template<typename T>
ostream& operator+(ostream& out, const vector<T> &vec){
    for(const auto &x : vec){
        out<<x<<" ";
    }
    out<<"\n";
    return out;
}
template<typename T>
ostream& operator*(ostream& out, const vector<T> &vec){
    for(const auto &x : vec){
        out+x;
    }
    return out;
}
template<typename T>
istream& operator>>(istream& in, vector<T> &vec){
    for(auto &x : vec){
        in>>x;
    }
    return in;
}
void solve(){
    int n;
    cin>>n;
    vector<int> seq(n);
    cin>>seq;
    set<int> free;
    for(int i=2;i<=n;i++){
        free.insert(i);
    }
    function<bool(vector<int>&, set<int>&)> dfs = [&](vector<int> &curr, set<int> &left){
        if(left.size()==0){
            return true;
        }
        for(auto it = left.begin();it!=left.end();it++){
            int val = *it;
            if(val == seq[curr.back()-1]){
                continue;
            }
            curr.pb(val);
            free.erase(prev(++it));
            if(!dfs(curr, free)){
                curr.pop_back();
                free.insert(val);
                --it;
            }
            else{
                return true;
            }
        }
        return false;
    };
    vector<int> ans = {1};
    dfs(ans, free);
    if(ans.size()!=n){
        cout<<"-1\n";
        return;
    }
    cout+ans;
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
}