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>

#define ff first
#define ss second
#define pb push_back

using namespace std;
using ll = long long;
using ii = pair<int, int>;

const int N = 100005;

int cnt[N];
int v[N];

bool ok(vector<int> &ans){
	for(int i = 1; i < (int)ans.size(); i++){
		if(v[ans[i-1]] == ans[i]) return 0;
	}
	return 1;
}

int main() {
	int n;
	scanf("%d", &n);
	set<int> s;
	for(int i = 1; i <= n; i++){
		s.insert(i);
		scanf("%d", v+i);
		cnt[v[i]]++;
	}
	if(n == 2) return printf("-1\n"), 0;
	vector<int> ans;
	int lt = 0;
	for(int i = 0; i < n-3; i++){
		int p = *s.begin();
		if(p == lt){
			if(s.size() == 1){
				printf("-1\n");
				return 0;
			}
			p = *next(s.begin());
		}
		if(s.size()-1 && max(cnt[v[*s.begin()]], cnt[v[*s.rbegin()]]) == s.size()-1){
			int x = *s.begin();
			int y = *s.rbegin();
			p = cnt[v[x]] > cnt[v[y]] ? v[x] : v[y];
		}
		s.erase(p);
		ans.pb(p);
		lt = v[p];
		cnt[lt]--;
		cnt[lt] = max(cnt[lt], 0);
		cnt[p] = 0;
	}
	vector<int> v2;
	for(int i : s) v2.pb(i);
	do{
		for(int i : v2){
			ans.pb(i);
		}
		if(ok(ans)){
			break;
		}
		for(int i : v2){
			ans.pop_back();
		}
	}while(next_permutation(v2.begin(), v2.end()));
	for(int i : ans){
		printf("%d ", i);
	}
	printf("\n");

	return 0;
}
