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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <deque>
#include <set>
#include <iomanip>
#include <utility>
#define ll long long
#define ld long double
using namespace std;

int a[100010], num[100010], ans[100010];

int main() {
	int N;
	cin >> N;
	for(int i=1; i<=N; ++i){
		cin >> a[i];
		++num[a[i]];
	}
	if(N==2){
		cout << -1 << endl;
		return 0;
	}
	vector<set<int>> race(N);
	for(int i=1; i<=N; ++i) race[num[i]].insert(i);
	set<int> st;
	for(int i=1; i<=N; ++i) st.insert(i);
	int proh=0;
	for(int i=1; i+3<=N; ++i){
		if(!race[N-i].empty()){
			ans[i]=*race[N-i].begin();
		}else{
			auto itr=st.begin();
			if(*itr==proh) ++itr;
			ans[i]=*itr;
		}
		st.erase(ans[i]);
		race[num[ans[i]]].erase(ans[i]);
		proh=a[ans[i]];
		if(race[num[a[ans[i]]]].erase(a[ans[i]])==1){
			--num[a[ans[i]]];
			race[num[a[ans[i]]]].insert(a[ans[i]]);
		}
	}
	int cand[3], t[3];
	for(int i=0; i<3; ++i){
		cand[i]=*st.begin();
		st.erase(cand[i]);
	}
	if(proh!=cand[0] && a[cand[0]]!=cand[1] && a[cand[1]]!=cand[2]){
		t[0]=0, t[1]=1, t[2]=2;
	}else if(proh!=cand[0] && a[cand[0]]!=cand[2] && a[cand[2]]!=cand[1]){
		t[0]=0, t[1]=2, t[2]=1;
	}else if(proh!=cand[1] && a[cand[1]]!=cand[0] && a[cand[0]]!=cand[2]){
		t[0]=1, t[1]=0, t[2]=2;
	}else if(proh!=cand[1] && a[cand[1]]!=cand[2] && a[cand[2]]!=cand[0]){
		t[0]=1, t[1]=2, t[2]=0;
	}else if(a[cand[2]]!=cand[0]){
		t[0]=2, t[1]=0, t[2]=1;
	}else{
		t[0]=2, t[1]=1, t[2]=0;
	}
	for(int i=0; i<3; ++i) ans[N-2+i]=cand[t[i]];
	cout << ans[1];
	for(int i=2; i<=N; ++i) cout << ' ' << ans[i];
	cout << endl;
	return 0;
}