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;
#define sz(a) int(a.size())
typedef vector<int> vi;
#define pb push_back

int n,ans[100010],now[100010],a[100010],have[100010];
set<int> s,in[100010];

void chkmin() {
	for(int i=1;i<=n;i++)
		if(ans[i]!=now[i]) {
			if(now[i]<ans[i]) memcpy(ans,now,sizeof(ans));
			break;
		}
}

bool check() {
	for(int i=1;i<n;i++) if(a[now[i]]==now[i+1]) return 0;
	return 1;
}

void del(int x) {
	in[a[x]].erase(x);
}

void choose(int x,int i) {
	if(!have[a[x]]||sz(in[a[x]])!=n-i) now[i]=x;
	else now[i]=a[x];
}

int main() {
	ios::sync_with_stdio(0);cin.tie(0);
	cin>>n;ans[1]=n+1;
	for(int i=1;i<=n;i++) cin>>a[i],s.insert(i),in[a[i]].insert(i),have[i]=1;
	for(int i=1;i+4<=n;i++) {
		int x=*s.begin();
		if((i==1||a[now[i-1]]!=x)) choose(x,i);
		else s.erase(x),choose(*s.begin(),i),s.insert(x);
		del(now[i]);have[now[i]]=0;s.erase(now[i]);//cerr<<now[i]<<' ';
	}
	int lo=max(1,n-3),p=lo;
	while(!s.empty()) now[p++]=*s.begin(),s.erase(s.begin());
	do {
		if(check()) chkmin();
	} while(next_permutation(now+lo,now+n+1));
	if(ans[1]==n+1) cout<<-1;
	else for(int i=1;i<=n;i++) cout<<ans[i]<<' ';
	return 0;
}
