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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for(int i = (int)a; i <= (int)b; i++)
#define DEC(i, a, b) for(int i = (int)a; i >= (int)b; i--)
typedef pair<int, int> pi;
#define f first
#define s second
#define pb push_back
#define fastio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

int n, arr[100005], coun[100005];
set<pi> s;
set<int> pos;

int main() {
	fastio; cin >> n;
	if (n == 2) {cout << -1; return 0;}
	FOR(i, 1, n) {
		cin >> arr[i];
		coun[arr[i]]++;
	}
	FOR(i, 1, n) {
		s.insert(pi(coun[i], i));
		pos.insert(i);
	}
	int no = -1;
	while (n) {
		if (n == 3) {
			vector<int> a, v; a.clear(); v.clear();
			for(auto it:pos) a.pb(it);
			do {
				bool ok = true; int nn = no;
				for(auto it:a) {
					if (nn == it) {ok = false; break;}
					nn = arr[it];
				}
				if (ok) {
					if (v.empty()) v = a;
					else v = min(v, a);
				}
			} while (next_permutation(a.begin(), a.end()));
			for(auto it:v) cout << it << " ";
			return 0;
		}
		pi t = *(--s.end());
		int a = t.f, b = t.s;
		int cur = 0;
		if (a == n-1) cur = b;
		else {
			for (auto it:pos) {
				if (it != no) {
					cur = it;
					break;
				}
			}
		}
		cout << cur << " ";
		pos.erase(cur); s.erase(pi(coun[cur], cur));
		no = arr[cur]; n--;
		if (pos.find(no) != pos.end()) {
			s.erase(pi(coun[no], no));
			--coun[no];
			s.insert(pi(coun[no], no));
		}
	}
}