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 <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <math.h>
#include <set>
using namespace std;
typedef long long int ll;

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int n; cin >> n;
	vector<int> a(n);
	for(int i=0;i<n;i++){
		cin >> a[i];
		a[i]--;
	}
	vector<int> d(n);
	for(int i=0;i<n;i++){
		d[a[i]]++;
	}
	vector<int> ans(n,-1);
	int ban=-1;
	set<int> st;
	for(int i=0;i<n;i++){
		st.insert(i);
	}
	for(int i=0;i<n;i++){
		if(n-i<=8){
			vector<int> v;
			for(int vv:st){
				v.push_back(vv);
			}
			do{
				bool ok=1;
				if(v[0]==ban)ok=0;
				for(int j=0;j<v.size()-1;j++){
					if(a[v[j]]==v[j+1])ok=0;
				}
				if(ok){
					for(int j=0;j<v.size();j++){
						ans[i+j]=v[j];
					}
					for(int j=0;j<n;j++){
						cout << ans[j]+1 << " ";
					}
					cout << endl;
					return 0;
				}
			}while(next_permutation(v.begin(),v.end()));
			cout << -1 << endl;
			return 0;
		}
		else{
			int k=n-i;
			for(int v:st){
				if(v==ban)continue;
				if(st.count(a[v])&&d[a[v]]==k-1)continue;
				ans[i]=v;
				st.erase(v);
				ban=a[v];
				d[a[v]]--;
				break;
			}
		}
	}
}