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<cmath>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<set>
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
int stack[20];
inline void write(int x)
{
	if(x<0){putchar('-');x=-x;}
	if(!x){putchar('0');return;}
	int top=0;
	while(x)stack[++top]=x%10,x/=10;
	while(top)putchar(stack[top--]+'0');
}
inline void pr1(int x){write(x),putchar(' ');}
inline void pr2(int x){write(x),puts("");}
set<int>s;
typedef set<int>::iterator It;
int n,a[100010],cnt[100010];
int top,ans[100010];
inline bool dfs(int limit)
{
	if(top==n){for(int i=1;i<=n;i++)pr1(ans[i]);return true;}
	if(n-top-1==cnt[limit] && s.find(limit)!=s.end())return false;
	for(It i=s.begin();i!=s.end();i++)
	{
		int p=*i;
		if(p==limit)continue;
		ans[++top]=p,cnt[a[p]]--;s.erase(i);
		if(dfs(a[p])==true)return true;
		top--,cnt[a[p]]++;s.insert(p);
	}return false;
}
int main()
{
	//freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
	n=read();
	for(int i=1;i<=n;i++)a[i]=read(),cnt[a[i]]++,s.insert(i);
	if(dfs(0)==false)puts("-1");
	return 0;
}
