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 ll long long
typedef pair<int,int> pi;
#define f first
#define s second
#define FAST ios_base::sync_with_stdio(0); cin.tie(0);
#define all(x) x.begin(),x.end()

int n;
int A[100010],co[100010];
set <int> S;
//int ans[100010];

int32_t main() {
	cin >> n;
	for (int i =1;i<=n;i++) {
		cin >> A[i];
		co[A[i]]++;
		S.insert(i);
	}
	
	if (n == 1) {
		cout << "1";
		return 0;
	} else if (n == 2) {
		 if (A[1] != 2) cout << "12";
		 else if (A[2] != 1) cout << "21";
		 else cout << "-1";
		 return 0;
	}
	
	int prev = n+1;
	//int co = 0;
	while (S.size() > 3) {
		auto it = S.begin();
		//Conditions to fail:
		//1: Cannot put it after previous
		//2: Guys from i to n all have the same ban on x AND x has not been used yet. 
		while (A[prev] == *it or (S.find(A[*it]) != S.end() and  co[A[*it]] == (int)S.size()-1)) it++;
		
		cout << *it << " ";
		prev = *it;
		co[A[*it]]--;
		S.erase(it);
	}
	
	vector <int> v;
	for (auto it: S) {
		v.push_back(it);
	}
	
	//cout << v[0] << " " << v[1] << " " << v[2] << "\n";
	while (true) {
		if (A[prev] != v[0] and A[v[0]] != v[1] and A[v[1]] != v[2]) {
			cout << v[0] << " " << v[1] << " " << v[2];
			return 0;
		}
		next_permutation(v.begin(),v.end());
	}
}







