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)

#pragma region template
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vld = vector<ld>;
using vvld = vector<vld>;
using vvvld = vector<vvld>;
using vs = vector<string>;
using pll = pair<ll, ll>;
using vp = vector<pll>;
template <typename T>
using pqrev = priority_queue<T, vector<T>, greater<T>>;
#define rep(i, n) for (ll i = 0, i##_end = (n); i < i##_end; i++)
#define repb(i, n) for (ll i = (n)-1; i >= 0; i--)
#define repr(i, a, b) for (ll i = (a), i##_end = (b); i < i##_end; i++)
#define reprb(i, a, b) for (ll i = (b)-1, i##_end = (a); i >= i##_end; i--)
#define ALL(a) (a).begin(), (a).end()
#define SZ(x) ((ll)(x).size())
//*
constexpr ll MOD = 1e9 + 7;
/*/
constexpr ll MOD = 998244353;
//*/
constexpr ll INF = 1e+18;
constexpr ld EPS = 1e-12L;
constexpr ld PI = 3.14159265358979323846L;
constexpr ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
constexpr ll LCM(ll a, ll b) { return a / GCD(a, b) * b; }
template <typename S, typename T>
inline bool chmax(S &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename S, typename T>
inline bool chmin(S &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
#ifdef OJ_LOCAL
#include "dump.hpp"
#else
#define dump(...) ((void)0)
#endif
template <typename T>
bool print_(const T &a) {
  cout << a;
  return true;
}
template <typename T>
bool print_(const vector<T> &vec) {
  for (auto &a : vec) {
    cout << a;
    if (&a != &vec.back()) {
      cout << " ";
    }
  }
  return false;
}
template <typename T>
bool print_(const vector<vector<T>> &vv) {
  for (auto &v : vv) {
    for (auto &a : v) {
      cout << a;
      if (&a != &v.back()) {
        cout << " ";
      }
    }
    if (&v != &vv.back()) {
      cout << "\n";
    }
  }
  return false;
}
void print() { cout << "\n"; }
template <typename Head, typename... Tail>
void print(Head &&head, Tail &&... tail) {
  bool f = print_(head);
  if (sizeof...(tail) != 0) {
    cout << (f ? " " : "\n");
  }
  print(forward<Tail>(tail)...);
}
#pragma endregion

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  ll n;
  cin >> n;
  vll a(n);
  rep(i, n){
    cin >> a[i];
    a[i]--;
  }
  if(n <= 8){
    vi perm(n);
    iota(ALL(perm), 0);
    do{
      bool f = 1;
      repr(i, 1, n){
        if(perm[i] == a[perm[i-1]]){
          f = 0;
          break;
        }
      }
      if(f){
        rep(i, n) perm[i]++;
        print(perm);
        return 0;
      }
    }while(next_permutation(ALL(perm)));
    print(-1);
    return 0;
  }
  vll ans(n);
  set<int> cards;
  rep(i, n){
    cards.emplace(i);
  }
  rep(i, n-8){
    auto it = cards.begin();
    if(i > 0 && *it == a[ans[i-1]]){
      it++;
    }
    ans[i] = *it;
    cards.erase(it);
  }
  vi perm;
  for(auto&& e: cards){
    perm.emplace_back(e);
  }
  do{
    bool f = 1;
    if(perm[0] == a[ans[n-9]]){
      f = 0;
    }
    repr(i, 1, 8){
      if(perm[i] == a[perm[i-1]]){
        f = 0;
        break;
      }
    }
    if(f){
      rep(i, 8) ans[i+n-8] = perm[i];
      rep(i, n) ans[i]++;
      print(ans);
      return 0;
    }
  }while(next_permutation(ALL(perm)));
  return 1;
}