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: from sys import stdin
  input = stdin.readline
  
  from math import ceil
  
  def answer(a , b):
  
      if(a == 0 and b == 0):return [i for i in range(1 , n + 1)]
      if(n == 2 and (a or b)):return [-1]
  
      if(abs(a - b) > 1):return [-1]
      
      ans , front , back = [] ,1 , n
      if(a > b):
  
          ans.append(front)
          ans.append(back)
          front , back = front + 1 , back - 1
  
          turn = 1
          while(front <= back):
  
              if(b == 0):break
  
              if(turn == 1):
                  ans.append(front);a -= 1;front += 1
              else:
                  ans.append(back);b -= 1;back -= 1
  
              turn ^= 1
  
          if(front <= back):a -= 1
          while(front <= back):
              ans.append(back)
              back -= 1
  
      elif(a < b):
  
          ans.append(back)
          ans.append(front)
          front , back = front + 1 , back - 1
  
          turn = 1
          while(front <= back):
  
              if(a == 0):break
  
              if(turn == 1):
                  ans.append(back);b -= 1;back -= 1
              else:
                  ans.append(front);a -= 1;front += 1
  
              turn ^= 1
  
          if(front <= back):b -= 1
          while(front <= back):
              ans.append(front)
              front += 1
  
      else:
          ans.append(front)
          ans.append(back)
          front , back = front + 1 , back - 1
  
          turn = 1
          while(front <= back):
              if(a == 0 and b == 1):break
  
              if(turn == 1):
                  ans.append(front);a -= 1;front += 1
              else:
                  ans.append(back);b -= 1;back -= 1
  
              turn ^= 1
  
          if(front <= back):b -= 1
          while(front <= back):
              ans.append(front)
              front += 1
          
  
      if(a or b):return  [-1]
      return ans
              
          
  for T in range(int(input())):
  
      n , a , b = map(int,input().split())
      
      
      print(*answer(a , b))
  