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
n=int(input())
a=list(map(lambda x:int(x)-1,input().split()))

ans=[-1]*n

def calc1(a):
  m=len(a)
  b=[0]*m
  for i in range(m):
    b[(2*i)%m]=a[i]
  for i in range(m):
    ans[b[i]]=b[(i+1)%m]

def calc2(a,b):
  m=len(a)
  c=[0]*2*m
  for i in range(m):
    c[2*i]=a[i]
    c[2*i+1]=b[i]
  for i in range(2*m):
    ans[c[i]]=c[(i+1)%(2*m)]

memo=[[] for i in range(n+1)]
seen=[0]*n
for i in range(n):
  if seen[i]:
    continue
  loop=[i]
  seen[i]=1
  now=a[i]
  while now!=i:
    loop.append(now)
    seen[now]=1
    now=a[now]
  if len(loop)%2==1:
    calc1(loop)
  else:
    m=len(loop)
    if memo[m]:
      calc2(memo[m],loop)
      memo[m]=[]
    else:
      memo[m]=loop

if min(ans)!=-1:
  print(*[i+1 for i in ans])
else:
  print(-1)