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 N=1e5+5;
int n,a[N],i,x,cnt[N],pp,p[N];
set<int>s,ss[N];
inline void bf(){
	int p[9],xb=0;
	p[0]=::p[pp];for(int x:s)p[++xb]=x;
	do{
		for(i=0;i<xb;++i)if(a[p[i]]==p[i+1])break;if(i==xb)break;
	}while(next_permutation(p+1,p+xb+1));
	for(i=1;i<=xb;++i)::p[++pp]=p[i];
}
int main(){
	scanf("%d",&n);for(i=1;i<=n;++i)scanf("%d",a+i),++cnt[a[i]],s.insert(i);
	if(n==2){puts("-1");return 0;}
	for(i=1;i<=n;++i)ss[cnt[i]].insert(i);
	for(;;){
		if(n-pp<=5)
			{bf();break;}
		if(!ss[n-pp-1].empty())x=*ss[n-pp-1].begin();
			else x=*s.begin()==a[p[pp]]?*++s.begin():*s.begin();
		p[++pp]=x;s.erase(x);ss[cnt[x]].erase(x);
		if(s.count(a[x]))ss[cnt[a[x]]].erase(a[x]),ss[--cnt[a[x]]].insert(a[x]);
	}
	for(i=1;i<=n;++i)printf("%d%c",p[i],i==n?'\n':' ');
}