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
#define _USE_MATH_DEFINES

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <unordered_map> 
#include <unordered_set>
#include <functional>
#include <bitset>

using namespace std;

typedef pair<long long int, long long int> P;
typedef tuple<int, int, int> T;

long long int INF = 1e18;
long long int MOD = 1e9 + 7;

long long int a[110000], cnt[110000];

int main(){
	int N;
	cin >> N;
	if(N == 2){
		cout << -1 << endl;
		return 0;
	}
	map<int, int> m;
	for(int i = 1; i <= N; i++){
		cin >> a[i];
		cnt[a[i]] += 1;
		m[i] = 0;
	}
	int nex = 0;
	vector<int> v;
	for(int loop = 0; loop < N; loop++){
		auto ite = m.begin();
		while(true){
			int num = ite->first;
			//cout << loop << " " << ite->first << " hoge" << endl;
			if(num == nex){
				ite++;
				continue;
			}
			int rem = m.size() - 1;
			if(rem == 0){
				v.push_back(num);
				break;
			}
			if(cnt[a[num]] >= rem){
				ite++;
				continue;
			}
			if(rem == 2){
				vector<int> v;
				for(P p : m){
					if(num != p.first){
						v.push_back(p.first);
					}
				}
				if(a[v[0]] == v[1] && a[v[1]] == v[0]){
					ite++;
					continue;
				}
			}
			cnt[a[num]] -= 1;
			cnt[num] = 0;
			m.erase(ite->first);
			nex = a[num];
			v.push_back(num);
			break;
		}
	}
	for(int i = 0; i < v.size() - 1; i++){
		cout << v[i] << " ";
	}
	cout << v[v.size() - 1] << endl;
	return 0;
}