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: for _ in range(int(input())):
      n,a,b=map(int,input().split())
      if(a+b+2<=n and abs(a-b)<=1):
          ans=[]
  
          for i in range(1,max(a,b)*2+3,):
              ans.append(i%2)
          for i in range(1,max(a,b)*2+3,2):
              ans[i-1]+=i 
              ans[i]+=i
          #ans=ans[:n]
          if(a<b):
              ans=ans[:len(ans)-2]+ans[len(ans)-2:][::-1]
              ans=ans[:n]
          elif(b<a):
              ans=ans[:2][::-1]+ans[2:]
              ans=ans[1:]
              ans=[i-1 for i in ans]
              ans=ans[:n]
          ind=ans.index(max(ans))
          xxx=max(ans)
          x=[i+xxx for i in range(1,n-len(ans)+1)]
          ans=ans[:ind+1]+x+ans[ind+1:]
          if(max(a,b)==0):
              print(*[i for i in range(1,n+1)])
          else:
              ans=ans[:n]
              print(*ans)
      elif(max(a,b)==0):
          print(*[i for i in range(1,n+1)])
      else:
          print(-1)