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: Runtime Error)

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <math.h>
#include <iomanip>
#include <limits>
#include <list>
#include <queue>
#include <tuple>
#include <map>
#include <stack>
#include <set>
using namespace std;
#define fast_io ios_base::sync_with_stdio (false) ; cin.tie(0) ; cout.tie(0) ;
#define MOD (long long int)(1e9+7)
#define ll long long int
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define reps(i,n) for(int i=1; i<=(int)(n); i++)
#define REP(i,n) for(int i=n-1; i>=0; i--)
#define REPS(i,n) for(int i=n; i>0; i--)
#define INF (int)(1123456789)
#define LINF (long long int)(112345678901234567)
#define chmax(a, b) a = (((a)<(b)) ? (b) : (a))
#define chmin(a, b) a = (((a)>(b)) ? (b) : (a))
#define all(v) v.begin(), v.end()

typedef pair<int, int> Pii;
typedef pair<ll, ll> Pll;

const int N = (int)3e5;

ll mpow(ll a, ll b){
  if(b==0) return 1;
  else if(b%2==0){ll memo = mpow(a,b/2); return memo*memo%MOD;}
  else return mpow(a,b-1) * a % MOD;
}
ll lpow(ll a, ll b){
  if(b==0) return 1;
  else if(b%2==0){ll memo = lpow(a,b/2); return memo*memo;}
  else return lpow(a,b-1) * a;
}
ll gcd(ll a, ll b){
  if(b==0) return a;
  else return gcd(b, a%b);
}
vector<ll> kaijo_memo;
ll kaijo(ll n){
  if(kaijo_memo.size() > n) return kaijo_memo[n];
  if(kaijo_memo.size() == 0) kaijo_memo.push_back(1);
  while(kaijo_memo.size() <= n) kaijo_memo.push_back(kaijo_memo[kaijo_memo.size()-1] * kaijo_memo.size() % MOD);
  return kaijo_memo[n];
}
vector<ll> gyaku_kaijo_memo;
ll gyaku_kaijo(ll n){
  if(gyaku_kaijo_memo.size() > n) return gyaku_kaijo_memo[n];
  if(gyaku_kaijo_memo.size() == 0) gyaku_kaijo_memo.push_back(1);
  while(gyaku_kaijo_memo.size() <= n) gyaku_kaijo_memo.push_back(gyaku_kaijo_memo[gyaku_kaijo_memo.size()-1] * mpow(gyaku_kaijo_memo.size(), MOD-2) % MOD);
  return gyaku_kaijo_memo[n];
}

ll nCr(ll n, ll r){
  if(n == r) return 1;//0個の丸と-1個の棒みたいな時に時に効く？不安.
  if(n < r || r < 0) return 0;
  ll ret = 1;
  ret *= kaijo(n); ret %= MOD;
  ret *= gyaku_kaijo(r); ret %= MOD;
  ret *= gyaku_kaijo(n-r); ret %= MOD;
  return ret;
}

int main(void){
  fast_io
  ll n;cin>>n;
  vector<ll> A(n);
  vector<bool> used(n);
  map<ll,ll> mp;
  rep(i,n){
    cin>>A[i];A[i]--;
    if(mp.count(A[i]) == 0){
      mp[A[i]] = 0;
    }
    mp[A[i]]++;
    used[i] = false;
  }
  vector<ll> ans;
  int max = -1;
  int memo = -1;
  int remain = n;
  if(n==2){
    if(A[0] != 1){
      cout<<"1 2"<<endl;
      return 0;
    }
    if(A[0] == 1){
      if(A[1] != 0){
        cout<<"2 1"<<endl;
      }else{
        cout<<-1<<endl;
      }
      return 0;
    }
  }
  while(!(max == n-1 && memo == -1)){
    if(remain == 3){
      int bad = -10;
      if(ans.size() > 0){
        bad = A[ans.back()];
      }
      vector<ll> nokori;
      rep(i,n){
        if(!used[i]){
          nokori.push_back(i);
        }
      }
      if(bad != nokori[0] && A[nokori[0]] != nokori[1] && A[nokori[1]] != nokori[2]){
        ans.push_back(nokori[0]);
        ans.push_back(nokori[1]);
        ans.push_back(nokori[2]);
        break;
      }
      if(bad != nokori[0] && A[nokori[0]] != nokori[2] && A[nokori[2]] != nokori[1]){
        ans.push_back(nokori[0]);
        ans.push_back(nokori[2]);
        ans.push_back(nokori[1]);
        break;
      }
      if(bad != nokori[1] && A[nokori[1]] != nokori[0] && A[nokori[0]] != nokori[2]){
        ans.push_back(nokori[1]);
        ans.push_back(nokori[0]);
        ans.push_back(nokori[2]);
        break;
      }
      if(bad != nokori[1] && A[nokori[1]] != nokori[2] && A[nokori[2]] != nokori[0]){
        ans.push_back(nokori[1]);
        ans.push_back(nokori[2]);
        ans.push_back(nokori[0]);
        break;
      }
      if(bad != nokori[2] && A[nokori[2]] != nokori[0] && A[nokori[0]] != nokori[1]){
        ans.push_back(nokori[2]);
        ans.push_back(nokori[0]);
        ans.push_back(nokori[1]);
        break;
      }
      if(bad != nokori[2] && A[nokori[2]] != nokori[1] && A[nokori[1]] != nokori[0]){
        ans.push_back(nokori[2]);
        ans.push_back(nokori[1]);
        ans.push_back(nokori[0]);
        break;
      }
    }
    if(mp.size() == 2){
      for(auto itr = mp.begin(); itr != mp.end(); itr++){
        ll key = itr->first;
        ll value = itr->second;
        if(value == remain-1){//いじめられてる
          ans.push_back(key);
          used[key] = true;
          vector<ll> nokori;
          rep(i,n){
            if(!used[i]){
              nokori.push_back(i);
            }
          }
          if(nokori[0] != A[key]){
            rep(i,nokori.size()){
              ans.push_back(nokori[i]);
            }
          }else{
            ans.push_back(nokori[1]);
            rep(i,nokori.size()){
              if(i==1) continue;
              ans.push_back(nokori[i]);
            }
          }

          rep(i,ans.size()){
            cout<<ans[i]+1<<" ";
          }
          cout<<endl;
          return 0;

        }
      }
    }
    int bad = -10;
    if(ans.size() > 0){
      bad = A[ans.back()];
    }
    if(memo != -1 && bad != memo){
      ans.push_back(memo);
      used[memo] = true;
      memo = -1;
      continue;
    }
    if(bad == max+1){
      memo = max+1;
      ans.push_back(max+2);
      used[max+2] = true;
      max += 2;
      continue;
    }else{
      ans.push_back(max+1);
      used[max+1] = true;
      max += 1;
    }
  }
  rep(i,ans.size()){
    cout<<ans[i]+1<<" ";
  }
  cout<<endl;
  return 0;
}