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: t = int(input())
  
  for i in range(t) :
      n, a , b  = input().split()
      n = int(n)
      a= int(a)
      b= int(b)
      l = []
      kaf = int ((n-1)/2 )
      if a > kaf or b> kaf or a-b > 1 or b-a > 1  or a + b > n -2:
          print(-1)
      else :
          if( a> b) :
              i = 0
              p = 1
              l.append(p)
              p += 1
              q = n
              while ( i < b) :
  
                  l.append(q)
                  q -= 1
                  l.append(p)
                  p += 1
                  i += 1
              while (q >= p) :
                  l.append(q)
                  q -= 1
          elif a < b :
              i = 0
              p = 1
              q = n
  
              l.append(q)
              q -= 1
              while (i < a):
                  l.append(p)
                  p += 1
                  l.append(q)
                  q -= 1
                  i += 1
              while (p<= q):
                  l.append(p)
                  p += 1
          else:
              i = 0
              p = 1
              l.append(p)
              p += 1
              q = n
              while (i < b):
                  l.append(q)
                  q -= 1
                  l.append(p)
                  p += 1
                  i += 1
              while (p <= q):
                  l.append(p)
                  p += 1
  
  
      print(*l)
  
  
  
  # p = "easy"
  #
  # l = list(map(int,input().strip().split()))[:n]
  # for i in range (n) :
  #     if l[i] == 1 :
  #         p = "hard"
  # print(p)
  # l = list(map(int,input().strip().split()))[:n]
  # t= []
  # st = "HQ9"
  # l = list(st)
  # print(l)
  # sum = 0
  # st= "1111"
  # p = 0
  # s = input()
  # i = 0
  # while i < n :
      # t.append(l.index(i+1)+ 1)
      # s= input()
      # if st != s :
      #     sum += 1
      #     st = s
      # i += 1
      # for i in range(n):
  # print(*t)
      # a, b = input().split()
      # a = int(a)
      # b = int(b)
      # if b-a > 1 :
      #     sum += 1
      # if l.count(s[i]) > 0 :
  #         p =1
  #     i += 1
  #
  # if p == 1:
  #     print("YES")
  # else :
  #     print("NO")
  # print(sum)
  