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<cstdio>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
using namespace std;
int a[100005],n,vis[100005],tb,tmp;
set<int>al;
map<int,int>mp;
map<int,int>::iterator it,it2; 
set<int>::iterator its;
void sel(int p)
{
	printf("%d ",p);
	al.erase(p);
	vis[p]=0;
	if(mp.find(p)!=mp.end())mp.erase(p);
	it=mp.find(a[p]);
	if(it!=mp.end())
	{
		it->second--;
		if(it->second==0)mp.erase(it);
	}
	tb=a[p];
}
bool chk(int p)
{
	if((int)mp.size()>2)return false;
	for(it=mp.begin();it!=mp.end();++it)if(it->second==p)return true;
	return false;
}
int main()
{
	scanf("%d",&n);
	if(n==2)
	{
		printf("-1\n");
		return 0;
	}
	for(int i=1;i<=n;i++)vis[i]=1,al.insert(i);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),mp[a[i]]++;		 
	tb=0;
	for(int i=1;i<=n;i++)
	{
		if(chk(n-i))
		{
			for(it=mp.begin();it!=mp.end();it++)
			{
				if(it->second==n-i)
				{
					sel(it->first);
					break;
				}
			}
			continue;
		}
		if(i==n-2&&(int)mp.size()==2&&a[a[mp.begin()->first]]==mp.begin()->first)
		{
			it2=it=mp.begin();
			it2++;
			if(tb==it->first)sel(it2->first);
			else if(tb==it2->first||it->first<it2->first)sel(it->first);
			else sel(it2->first);
			continue;
		}
		its=al.begin();
		if(*its==tb)its++;
		sel(*its);
	}
	return 0;
}