description: The city where Mocha lives in is called Zhijiang. There are n+1 villages and 2n-1 directed roads in this city. 
  
  There are two kinds of roads:
  
    * n-1 roads are from village i to village i+1, for all 1≤ i ≤ n-1. 
    * n roads can be described by a sequence a_1,…,a_n. If a_i=0, the i-th of these roads goes from village i to village n+1, otherwise it goes from village n+1 to village i, for all 1≤ i≤ n. 
  
  
  
  Mocha plans to go hiking with Taki this weekend. To avoid the trip being boring, they plan to go through every village exactly once. They can start and finish at any villages. Can you help them to draw up a plan? 
  
  Input
  
  Each test contains multiple test cases. 
  
  The first line contains a single integer t (1 ≤ t ≤ 20) — the number of test cases. Each test case consists of two lines.
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 10^4) — indicates that the number of villages is n+1.
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1). If a_i=0, it means that there is a road from village i to village n+1. If a_i=1, it means that there is a road from village n+1 to village i.
  
  It is guaranteed that the sum of n over all test cases does not exceed 10^4.
  
  Output
  
  For each test case, print a line with n+1 integers, where the i-th number is the i-th village they will go through. If the answer doesn't exist, print -1.
  
  If there are multiple correct answers, you can print any one of them.
  
  Example
  
  Input
  
  
  2
  3
  0 1 0
  3
  1 1 0
  
  
  Output
  
  
  1 4 2 3 
  4 1 2 3 
  
  Note
  
  In the first test case, the city looks like the following graph:
  
  <image>
  
  So all possible answers are (1 → 4 → 2 → 3), (1 → 2 → 3 → 4).
  
  In the second test case, the city looks like the following graph:
  
  <image>
  
  So all possible answers are (4 → 1 → 2 → 3), (1 → 2 → 3 → 4), (3 → 4 → 1 → 2), (2 → 3 → 4 → 1).
solution: #!/usr/bin/env python3
  import sys
  import getpass  # not available on codechef
  import math, random
  import functools, itertools, collections, heapq, bisect
  from collections import Counter, defaultdict, deque
  input = sys.stdin.readline  # to read input quickly
  
  # available on Google, AtCoder Python3, not available on Codeforces
  # import numpy as np
  # import scipy
  
  M9 = 10**9 + 7  # 998244353
  yes, no = "YES", "NO"
  # d4 = [(1,0),(0,1),(-1,0),(0,-1)]
  # d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
  # d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)]  # hexagonal layout
  MAXINT = sys.maxsize
  
  # if testing locally, print to terminal with a different color
  OFFLINE_TEST = getpass.getuser() == "hkmac"
  # OFFLINE_TEST = False  # codechef does not allow getpass
  def log(*args):
      if OFFLINE_TEST:
          print('\033[36m', *args, '\033[0m', file=sys.stderr)
  
  def solve(*args):
      # screen input
      if OFFLINE_TEST:
          log("----- solving ------")
          log(*args)
          log("----- ------- ------")
      return solve_(*args)
  
  def read_matrix(rows):
      return [list(map(int,input().split())) for _ in range(rows)]
  
  def read_strings(rows):
      return [input().strip() for _ in range(rows)]
  
  def minus_one(arr):
      return [x-1 for x in arr]
  
  def minus_one_matrix(mrr):
      return [[x-1 for x in row] for row in mrr]
  
  # ---------------------------- template ends here ----------------------------
  
  
  def solve_(lst, k):
      # your solution here
  
      if lst[0] == 1:
          return [k+1] + list(range(1,k+1))
  
      if lst[-1] == 0:
          return list(range(1,k+2))
  
      for i,(a,b) in enumerate(zip(lst, lst[1:]), start=1):
          if a == 0 and b == 1:
              return list(range(1,i+1)) + [k+1] + list(range(i+1,k+1))
  
      return [-1]
  
  
  
  # for case_num in [0]:  # no loop over test case
  # for case_num in range(100):  # if the number of test cases is specified
  for case_num in range(int(input())):
  
      # read line as an integer
      k = int(input())
  
      # read line as a string
      # srr = input().strip()
  
      # read one line and parse each word as a string
      # lst = input().split()
      
      # read one line and parse each word as an integer
      # a,b,c = list(map(int,input().split()))
      lst = list(map(int,input().split()))
      # lst = minus_one(lst)
  
      # read multiple rows
      # arr = read_strings(k)  # and return as a list of str
      # mrr = read_matrix(k)  # and return as a list of list of int
      # mrr = minus_one_matrix(mrr)
  
      res = solve(lst, k)  # include input here
  
      # print length if applicable
      # print(len(res))
  
      # parse result
      res = " ".join(str(x) for x in res)
      # res = "\n".join(str(x) for x in res)
      # res = "\n".join(" ".join(str(x) for x in row) for row in res)
  
      # print result
      # print("Case #{}: {}".format(case_num+1, res))   # Google and Facebook - case number required
  
      print(res)