A permutation of length *n* is an array containing each integer from 1 to *n* exactly once. For example, *q* = [4, 5, 1, 2, 3] is a permutation. For the permutation *q* the square of permutation is the permutation *p* that *p*[*i*] = *q*[*q*[*i*]] for each *i* = 1... *n*. For example, the square of *q* = [4, 5, 1, 2, 3] is *p* = *q*<sup class="upper-index">2</sup> = [2, 3, 4, 5, 1].

This problem is about the inverse operation: given the permutation *p* you task is to find such permutation *q* that *q*<sup class="upper-index">2</sup> = *p*. If there are several such *q* find any of them.
The first line contains integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">6</sup>) — the number of elements in permutation *p*.

The second line contains *n* distinct integers *p*<sub class="lower-index">1</sub>, *p*<sub class="lower-index">2</sub>, ..., *p*<sub class="lower-index">*n*</sub> (1 ≤ *p*<sub class="lower-index">*i*</sub> ≤ *n*) — the elements of permutation *p*.
none
4
2 1 4 3
3 4 2 1
4
2 1 3 4
-1
5
2 3 4 5 1
4 5 1 2 3
1
1
1
1
1
1
10
8 2 10 3 4 6 1 7 9 5
-1
10
3 5 1 2 10 8 7 6 4 9
6 9 8 10 4 3 7 1 5 2
100
11 9 35 34 51 74 16 67 26 21 14 80 84 79 7 61 28 3 53 43 42 5 56 36 69 30 22 88 1 27 65 91 46 31 59 50 17 96 25 18 64 55 78 2 63 24 95 48 93 13 38 76 89 94 15 90 45 81 52 87 83 73 44 49 23 82 85 75 86 33 47 19 58 97 37 20 40 10 92 4 6 68 77 54 71 12 62 60 100 39 41 99 72 29 57 8 70 32 66 98
-1
100
94 22 24 99 58 97 20 29 67 30 38 64 77 50 15 44 92 88 39 42 25 70 2 76 84 6 37 49 17 71 31 19 26 79 10 35 65 63 32 95 5 8 52 27 83 18 53 93 13 81 48 68 54 82 34 60 87 23 16 86 55 40 61 45 28 7 74 41 14 91 3 72 33 11 98 89 90 69 78 36 80 59 56 21 43 1 75 46 47 12 96 73 57 51 4 85 9 100 66 62
78 52 95 76 96 49 53 59 77 100 64 11 9 48 15 17 44 46 32 54 84 68 43 4 21 28 73 6 16 62 31 39 65 86 98 75 33 45 19 3 91 82 2 92 63 88 7 50 97 93 14 22 20 42 60 55 80 85 29 34 56 71 83 38 26 47 90 70 51 41 40 72 37 12 35 99 67 94 1 87 57 8 61 25 23 79 36 18 66 74 5 27 81 69 24 58 13 10 89 30
import math
import sys
input = sys.stdin.readline

n = int(input())
a = [int(_) - 1 for _ in input().split()]
vis = [False] * n
cycles = [[] for _ in range(n + 1)]
for i in range(n):
    if vis[i]: continue
    cur = i
    cycle = []
    while not vis[cur]:
        vis[cur] = True
        cycle.append(cur)
        cur = a[cur]
    cycles[len(cycle)].append(cycle)
p = [0] * n
for i in range(n + 1):
    if i % 2 == 1:
        for j in cycles[i]:
            for k in range(i):
                p[j[(k + 2) % i]] = j[k]
    else:
        if len(cycles[i]) % 2 == 1:
            print(-1)
            exit(0)
        for j in range(0, len(cycles[i]), 2):
            for k in range(i):
                p[cycles[i][j][k]] = cycles[i][j + 1][k] 
                p[cycles[i][j + 1][k]] = cycles[i][j][(k + 1) % i]
print(' '.join(map(lambda i : str(i + 1), p)))
