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
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

int const N = 100000;
int a[N + 1], an[N + 1], fr[N + 1], n;
set<pair<int, int> > st;
set<int> lft;

void rm(int t){
	lft.erase(t);
	st.erase(make_pair(fr[a[t]], a[t]));
	--fr[a[t]];
	st.insert(make_pair(fr[a[t]], a[t]));
}

void ad(int t){
	lft.insert(t);
	st.erase(make_pair(fr[a[t]], a[t]));
	++fr[a[t]];
	st.insert(make_pair(fr[a[t]], a[t]));
}

bool go(int i = 1, int bad = 0){
	if (i == n){
		int t = *lft.begin();
		if (t == bad)return false;
		an[i] = t;
		return true;
	}
	int m = n - i + 1;
	int t = st.rbegin()->second;
	if (fr[t] == m - 1 && lft.find(t) != lft.end()){
		an[i] = t;
		rm(t);
		bool ok = go(i + 1, a[t]);
		ad(t);
		return ok;
	}
	vector<int> v;
	int z = 5;
	for (auto it = lft.begin(); it != lft.end() && z; ++it, --z)v.push_back(*it);
	for (int u: v)if (bad != u){
		rm(u);
		an[i] = u;
		bool ok = go(i + 1, a[u]);
		ad(u);
		if (ok)return true;
	}
	return false;
}

int main(){
	scanf("%d", &n);
	if (n == 2){
		printf("-1\n");
		return 0;
	}
	f(i, 1, n + 1)scanf("%d", a + i), ++fr[a[i]];
	f(i, 1, n + 1)st.insert(make_pair(fr[i], i));
	f(i, 1, n + 1)lft.insert(i);
	go();
	printf("%d", an[1]);
	f(i, 2, n + 1)printf(" %d", an[i]);
	printf("\n");
}