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 modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000


int main(){
	
	int N;
	cin>>N;
	
	vector<int> a(N);
	
	set<int> notused,used;
	map<int,int> mp;
	
	for(int i=0;i<N;i++){
		cin>>a[i];
		a[i]--;
		mp[a[i]]++;
		notused.insert(i);
	}
	
	vector<int> ans;
	
	for(int i=0;i<N;i++){
		
		if(notused.size()<=3){
			vector<int> K;
			for(auto x:notused){
				K.push_back(x);
			}
			do{
				bool f = true;
					//cout<<f<<endl;
				if(ans.size()!=0){
					if(a[ans.back()]==K[0])f=false;
				}
			
				for(int j=1;j<K.size();j++){
					if(a[K[j-1]] == K[j])f=false;
				}
				if(f){
					for(int j=0;j<K.size();j++)ans.push_back(K[j]);
					break;
				}
			}
			while(next_permutation(K.begin(),K.end()));		
			
			break;
		}
		
		
		auto it = notused.begin();
		if(i!=0){
			if((*it)==a[ans.back()]){
				if(notused.size()==1){
					cout<<-1<<endl;
					return 0;
				}
				it++;
			}
		}
		
		int g = (*it);
		//cout<<g<<endl;
		if(i!=N-1){
			if(mp[a[g]] == N-1-i&&notused.count(a[g])){
				g = a[g];
			}
		}
		
		ans.push_back(g);
		mp[a[g]] --;
		
		if(!notused.count(g)){
			cout<<-1<<endl;
			return 0;
		}

		notused.erase(g);
		
	}
	
	if(ans.size()!=N){
		cout<<-1<<endl;
		return 0;
	}
	
	for(int i=0;i<N-1;i++){
		if(a[ans[i]] == ans[i+1]){
			cout<<-1<<endl;
			return 0;
		}
	}
	
	for(int i=0;i<N;i++){
		if(i!=0)cout<<' ';
		cout<<ans[i]+1;
	}
	cout<<endl;
	
    return 0;
}
