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>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define pb push_back
using namespace std;
using ll = long long;
using vi = vector<ll>;
using pi = pair<ll, ll>;
const int maxn = 2e5 + 33, mod = 1e9 + 7;
#define left dsajfil
int n, to[maxn], deg[maxn];
set<pi, greater<pi>> bydeg;
set<int> left;
void erase(int x) {
	left.erase(x);
	bydeg.erase({deg[x], x});
	x = to[x];
	bydeg.erase({deg[x], x});
	bydeg.insert({--deg[x], x});
}
vi ans;
void brute() {
	vi t;
	for(auto i : left) t.pb(i);
	do {
		if((ans.empty()||to[ans.back()]!=t[0])&&to[t[0]]!=t[1]&&to[t[1]]!=t[2]) {
			for(auto i : t) ans.pb(i);
			return;
		}
	} while(next_permutation(all(t)));
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	memset(deg, 0, sizeof deg);
	cin >> n;
	if(n==2) return cout << -1, 0;
	for(int i = 1; i <= n; i++) cin >> to[i], deg[to[i]]++;
	for(int i = 1; i <= n; i++) {
		left.insert(i);
		bydeg.insert({deg[i], i});
	}
	while(true) {
		if(left.size()==3) {
			brute();
			break;
		}
		if(bydeg.begin()->first == left.size()-1) {
			ans.pb(bydeg.begin()->second);
			erase(bydeg.begin()->second);
		} else {
			auto it = left.begin();
			if(!ans.empty()&&to[ans.back()]==*it) it++;
			ans.pb(*it);
			erase(*it);
		}
	}
	for(auto i : ans) cout << i << " ";
}

