description: You are given three integers n, a, b. Determine if there exists a permutation p_1, p_2, …, p_n of integers from 1 to n, such that:
  
    * There are exactly a integers i with 2 ≤ i ≤ n-1 such that p_{i-1} < p_i > p_{i+1} (in other words, there are exactly a local maximums).
  
    * There are exactly b integers i with 2 ≤ i ≤ n-1 such that p_{i-1} > p_i < p_{i+1} (in other words, there are exactly b local minimums).
  
  
  
  
  If such permutations exist, find any such permutation.
  
  Input
  
  The first line of the input contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. The description of test cases follows.
  
  The only line of each test case contains three integers n, a and b (2 ≤ n ≤ 10^5, 0 ≤ a,b ≤ n).
  
  The sum of n over all test cases doesn't exceed 10^5.
  
  Output
  
  For each test case, if there is no permutation with the requested properties, output -1.
  
  Otherwise, print the permutation that you are found. If there are several such permutations, you may print any of them.
  
  Example
  
  Input
  
  
  3
  4 1 1
  6 1 2
  6 4 0
  
  
  Output
  
  
  1 3 2 4
  4 2 3 1 5 6
  -1
  
  Note
  
  In the first test case, one example of such permutations is [1, 3, 2, 4]. In it p_1 < p_2 > p_3, and 2 is the only such index, and p_2> p_3 < p_4, and 3 the only such index.
  
  One can show that there is no such permutation for the third test case.
solution: import sys,os,io
  import math 
  from collections import defaultdict
  def ii():
      return int(input())
  def li():
      return list(map(int,input().split()))
  if(os.path.exists('input.txt')):
      sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w") 
  else:
      input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  
  def solve():
      n,a,b = li()
      cnt = n - 2 
      if n==2:
          if a==b==0:
              print(1,2)
          else:
              print(-1)
          return 
      if a+b>cnt:
          print(-1)
          return 
      if abs(a-b)>1:
          print(-1)
          return 
      
      if b>=a:#local minimum more 
          ans = [i for i in range(1,n+1)]
          ans.reverse()
          ind = 1
          for i in range(b):
              ans[ind],ans[ind+1]=ans[ind+1],ans[ind]
              ind+=2 
          if b>a:
              x = ans[ind-1]
              y = ans[ind:]
              ans[ind-1:-1]=y 
              ans[-1]=x 
      else:
          ans = [i for i in range(1,n+1)]
          ans.reverse()
          ind = 1
          for i in range(a):
              ans[ind],ans[ind+1]=ans[ind+1],ans[ind]
              ind+=2 
          x = ans[ind-1]
          y = ans[ind:]
          ans[ind-1:-1]=y 
          ans[-1]=x 
          # print(*ans)
          for i in range(n):
              ans[i] = n - ans[i]+1
      print(*ans)
  
          
  
              
  
  
  t = 1
  t = ii()
  for _ in range(t):
      solve()
  