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
import sys
from itertools import permutations
n = int(input())
a = [0] + list(map(int, input().split()))
if n == 2:
	print(-1)
	sys.exit()
used = [False for _ in range(n+1)]
hate = dict()
for x in a[1:]:
	if x in hate:
		hate[x] += 1
	else:
		hate[x] = 1
cur = 1
ng = 0
ans = []
while len(ans) < n-3:
	if len(hate) == 2:
		worst = False
		for k, v in hate.items():
			if v == n - len(ans) - 1 and not used[k]:
				worst = k
				break
		if worst:
			ans.append(k)
			used[k] = True
			ng = a[k]
			hate[a[k]] -= 1
			if hate[a[k]] == 0:
				del hate[a[k]]
			continue

	while used[cur]:
		cur += 1

	if cur == ng:
		cur_copy = cur+1
		while used[cur_copy]:
			cur_copy += 1
		ans.append(cur_copy)
		used[cur_copy] = True
		ng = a[cur_copy]
		hate[ng] -= 1
		if hate[ng] == 0:
			del hate[ng]
	else:
		ans.append(cur)
		used[cur] = True
		ng = a[cur]
		hate[ng] -= 1
		if hate[ng] == 0:
			del hate[ng]

yet = [i for i in range(1, n+1) if not used[i]]
for p in permutations(yet):
	if p[0] != ng and p[1] != a[p[0]] and p[2] != a[p[1]]:
		ans += list(p)
		break

print(*ans)