# input
N = gets.to_i
A = N.times.map{
      gets.to_i
    }

# solve
if A.uniq.size == 1
  print -1
else
  kouho = (0..N-1).to_a
  if A.last == kouho.last
    tmp = kouho.first
    kouho[0] = kouho.last
    kouho[N-1] = tmp
  end
  result =
    A.map{|num|
      tmp = kouho.first == num ? kouho[1] : kouho[0]
      kouho.delete(tmp)
      tmp
    }
  result.each{|num|
    print "#{num}\n"
  }
end