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;
int n,m,i,j,a[100005],cnt[100005];
set<int> v;
vector<int> ans;
void print(){
	for(i=0;i<n;i++){
		printf("%d%c",ans[i],(i+1==n?'\n':' '));
	}
	exit(0);
}
void inc(int x){
	ans.push_back(x);
	v.erase(x);
}
void dec(){
	v.insert(ans.back());
	ans.pop_back();
}
void dfs(int ban){
	if(ans.size()==n){
		print();
	}
	if(v.count(ban)&&cnt[ban]==n-ans.size()){
		return;
	}
	cnt[ban]--;
	for(set<int>::iterator it=v.begin();it!=v.end();it++){
		if(*it==ban) continue;
		inc(*it);
		dfs(a[*it]);
		dec();
	}
	cnt[ban]++;
}
int main(){
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%d",&a[i]);
		cnt[a[i]]++;
		v.insert(i);
	}
	for(i=1;i<=n;i++){
		inc(i);
		dfs(a[i]);
		dec();
	}
	puts("-1");
	return 0;
}