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 <iostream>
#include <iomanip>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <stack>
#include <queue>
#include <bitset>
#include <numeric>
#include <cassert>
#include <array>
#include <memory>
#ifdef DEBUG
#include "./Lib/debug.hpp"
#include "./Lib/Timer.hpp"
#include "./Lib/sample.hpp"
#else
#define dump(...)
#endif

/* (=＾o＾=) */
#define int ll

/* macro */
#define FOR(i, b, e) for(ll i = (ll)(b); i < (ll)(e); ++i)
#define RFOR(i, b, e) for(ll i = (ll)(e-1); i >= (ll)(b); --i)
#define REP(i, n) FOR(i, 0, n)
#define RREP(i, n) RFOR(i, 0, n)
#define REPC(x,c) for(const auto& x:(c))
#define REPI2(it,b,e) for(auto it = (b); it != (e); ++it)
#define REPI(it,c) REPI2(it, (c).begin(), (c).end())
#define RREPI(it,c) REPI2(it, (c).rbegin(), (c).rend())
#define REPI_ERACE2(it, b, e) for(auto it = (b); it != (e);)
#define REPI_ERACE(it, c) REPI_ERACE2(it, (c).begin(), (c).end())
#define ALL(x) (x).begin(),(x).end()
#define cauto const auto&
/* macro func */
template<class T>
inline auto sort(T& t) { std::sort(ALL(t)); }
template<class T>
inline auto rsort(T& t) { std::sort((t).rbegin(), (t).rend()); }
template<class T>
inline auto unique(T& t) { (t).erase(unique((t).begin(), (t).end()), (t).end()); }
template<class T, class S>
inline auto chmax(T& t, const S& s) { if (s > t) { t = s; return true; } return false; }
template<class T, class S>
inline auto chmaxE(T& t, const S& s) { if (s >= t) { t = s; return true; } return false; }
template<class T, class S>
inline auto chmin(T& t, const S& s) { if (s < t) { t = s; return true; } return false; }
inline auto BR() { std::cout << "\n"; }

/* type define */
using ll = long long;
using PAIR = std::pair<ll, ll>;
using VS = std::vector<std::string>;
using VL = std::vector<long long>;
using VVL = std::vector<VL>;
using VVVL = std::vector<VVL>;
using VD = std::vector<double>;
template<class T>
using V = std::vector<T>;

/* using std */
using std::cout;
constexpr char endl = '\n';
using std::cin;
using std::pair;
using std::string;
using std::stack;
using std::queue;
using std::vector;
using std::list;
using std::map;
using std::unordered_map;
using std::multimap;
using std::unordered_multimap;
using std::set;
using std::unordered_set;
using std::unordered_multiset;
using std::multiset;
using std::bitset;
using std::priority_queue;
using std::tuple;

/* constant value */
constexpr ll MOD = 1000000007;
//constexpr ll MOD = 998244353;

/* Initial processing  */
struct Preprocessing { Preprocessing() { std::cin.tie(0); std::ios::sync_with_stdio(0); }; }_Preprocessing;

/* Remove the source of the bug */
inline signed pow(signed, signed) { assert(false); return -1; }

/* define hash */
namespace std {
template <>	class hash<std::pair<ll, ll>> { public:	size_t operator()(const std::pair<ll, ll>& x) const { return hash<ll>()(1000000000 * x.first + x.second); } };
}

/* input */
template<class T> std::istream& operator >> (std::istream& is, vector<T>& vec) { for (T& x : vec) is >> x; return is; }

//=============================================================================================


auto solve_k(ll n, const VL& v, ll f) {
	dump(f);
	bool ok = true;
	priority_queue<ll, VL, std::greater<ll>> q;
	FOR(i, 0, n)if (i != f) { q.emplace(i); }
	list<ll> ans{f};
	ll now = f;
	REP(_, n - 1) {
		ll t = q.top();
		if (v[now] == t) {
			if (q.size() == 1) {
				ok = false;
				break;
			}
			q.pop();
			ll t2 = q.top();
			now = t2;
			ans.emplace_back(t2);
			q.pop();
			q.emplace(t);
		} else {
			now = t;
			ans.emplace_back(now);
			q.pop();
		}
	}
	if (ok) { return ans; }
	ll t = q.top();
	if (v[t] != f) {
		ans.emplace_front(t);
		return ans;
	}
	return list<ll>();
}

auto solve(ll n, const VL& v) {
	REP(i, 10)if (i < n) {
		auto ans = solve_k(n, v, i);
		if (!ans.empty()) { return ans; }
	}
	return list<ll>();
}

signed main() {
	ll n;
	cin >> n;
	VL v(n);
	cin >> v;
	for (auto&& x : v) { --x; }
	auto ans = solve(n, v);
	if (ans.empty()) { cout << -1 << endl; return 0; }
	for (auto&& x : ans) { ++x; }
	REPC(x, ans) { cout << x << " "; }BR();
}