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;

const int _ = 1e5 + 7;
int N , nxt[_] , du[_] , pre; set < int > in;
set < pair < int , int > > mxdu;
void erase(int t){in.erase(t); mxdu.erase(make_pair(du[t] , t));}

int main(){
	cin >> N; for(int i = 1 ; i <= N ; ++i){cin >> nxt[i]; ++du[nxt[i]];}
	if(N == 2){puts("-1"); return 0;}
	for(int i = 1 ; i <= N ; ++i){in.insert(i); mxdu.insert(make_pair(du[i] , i));}
	for(int i = 1 ; i <= N - 3 ; ++i){
		if((--mxdu.end())->first == N - i) erase(pre = (--mxdu.end())->second);
		else{
			auto t = in.begin(); if(nxt[pre] == *t) ++t;
			erase(pre = *t); --du[nxt[pre]];
			if(in.find(nxt[pre]) != in.end()){
				mxdu.erase(make_pair(du[nxt[pre]] + 1 , nxt[pre]));
				mxdu.insert(make_pair(du[nxt[pre]] , nxt[pre]));
			}
		}
		printf("%d " , pre);
	}
	int arr[3]; auto t = in.begin(); for(int i = 0 ; i < 3 ; ++i , ++t) arr[i] = *t;
	do{
		if(nxt[pre] != arr[0] && nxt[arr[0]] != arr[1] && nxt[arr[1]] != arr[2]){
			for(int i = 0 ; i < 3 ; ++i) printf("%d " , arr[i]);
			return 0;
		}
	}while(next_permutation(arr , arr + 3));
	return 0;
}
