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 math
  
  t = int(input())
  
  for _ in range(t):
      n, a, b = map(int, input().split())
      if (abs(a-b) > 1 or (a+b) > (n-2)):
          print(*[-1])
          continue
      elif (a == b == 0):
          print(*[i for i in range(1, n + 1)])
          continue
      lst = []
      max_a = 0
      min_b = 0
      if (a < b):
          lst.append(n)
          lst.append(1)
          cur = 2
          while(min_b != b - 1):
              lst.append(n - cur + 1)
              lst.append(cur)
              cur += 1
              min_b += 1
          for i in range(cur, n - cur + 1 + 1):
              lst.append(i)
      elif (a > b):
          lst.append(1)
          lst.append(n)
          cur = 2
          while (max_a != a - 1):
              lst.append(cur)
              lst.append(n - cur + 1)
              cur += 1
              max_a += 1
          for i in range(n - cur + 1, cur - 1, -1):
              lst.append(i)
      else:
          lst.append(1)
          lst.append(n)
          cur = 2
          fl = True
          while (min_b != b):
              if (fl):
                  lst.append(cur)
                  min_b += 1
                  fl = False
              else:
                  lst.append(n - cur + 1)
                  fl = True
              if fl == True:
                  cur += 1
          cur += 1
          for i in range(cur, n - cur + 1 + 1 + 1):
              lst.append(i)
      print(*lst)
  
  
  
  