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.

Examples

Input

4
2 3 4 1


Output

1 3 2 4


Input

2
2 1


Output

-1


Input

13
2 3 4 5 6 7 8 9 10 11 12 13 12


Output

1 3 2 4 6 5 7 9 8 10 12 11 13
#include <bits/stdc++.h>

using namespace std;
#define int long long
const int MOD = 1000000007;

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<int> A(N);
	map<int, int> mp;
	for (int i = 0; i < N; i++) {
		cin >> A[i];
		A[i]--;
		mp[A[i]]++;
	}
	if (N == 2) {
		cout << -1 << endl;
		return 0;
	}
	set<int> st;
	for (int i = 0; i < N; i++) {
		st.insert(i);
	}
	vector<int> res(N);
	int dame = -1;
	for (int i = 0; i < N - 3; i++) {
		auto a = st.begin();
		while ((*a) == dame){
				a++;
			}
		res[i] = (*a);
		dame = A[*a];
		mp[A[*a]]--;
		if (mp[A[*a]] == 0) {
			mp.erase(A[*a]);
		}
		st.erase(a);
	}
	vector<int> X;
	for (int i : st) {
		X.push_back(i);
	}
	bool ok = false;
	do {
		if (X[0] != dame && X[1] != A[X[0]] && X[2] != A[X[1]]) {
			ok = true;
			break;
		}
	} while (next_permutation(X.begin(), X.end()));
	for (int i = 0; i < 3; i++) {
		res[N - 3 + i] = X[i];
	}
	if (!ok) {
		int x = dame;
		vector<int> t;
		for (int i = N - 1; i >= -1; i--) {
			if (i == -1 || res[i] != dame) {
				if (i == -1 || A[res[i]] != dame) {
					sort(t.begin(), t.end());
					res[i + 1] = dame;
					if(A[dame] == t[0]){
						swap(t[0], t[1]);
					}
					for (int j = 0; j < t.size(); j++) {
						res[i + 2 + j] = t[j];
					}
					
					break;
				}
				t.push_back(res[i]);
			}
		}
	}
	for (int i = 0; i < N; i++) {
		if (i > 0) cout << " ";
		cout << res[i] + 1;
	}
	cout << endl;
}