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 itertools import *
  #from math import *
  #from bisect import *
  #from collections import *
  #from random import *
  #from decimal import *
  #from heapq import *
  #from itertools import *            # Things Change ....remember :)
  import sys
  input=sys.stdin.readline
  def inp():
      return int(input())
  def st():
      return input().rstrip('\n')
  def lis():
      return list(map(int,input().split()))
  def ma():
      return map(int,input().split())
  t=inp()
  while(t):
      t-=1
      n,a,b=ma()
      have=(n-2)//2
      have+=((n-2)%2)
      if(abs(a-b)>1):
          print(-1)
      else:
          if(a>have or b>have or (a+b)> n-2):
              print(-1)
          else:
              if(a>b):
                  vis=[0]*(n+1)
                  res=[]
                  lf=1
                  rf=n
                  for i in range(n):
                      if(a==0):
                          break
                      if(i%2==0):
                          res.append(lf)
                          vis[lf]=1
                          lf+=1
                      else:
                          res.append(rf)
                          vis[rf]=1
                          rf-=1
                          a-=1
                  for i in range(n,0,-1):
                      if(vis[i]==0):
                          res.append(i)
              elif(b>a):
                  lf=1
                  rf=n
                  vis=[0]*(n+1)
                  res=[]
                  for i in range(n):
                      if(b==0):
                          break
                      if(i%2):
                          res.append(lf)
                          vis[lf]=1
                          lf+=1
                          b-=1
                      else:
                          res.append(rf)
                          vis[rf]=1
                          rf-=1
                  for i in range(1,n+1):
                      if(vis[i]==0):
                          res.append(i)
              else:
                  vis=[0]*(n+1)
                  res=[]
                  lf=1
                  rf=n
                  for i in range(n):
                      if(a==0 and b==-1):
                          break
                      if(i%2==0):
                          res.append(lf)
                          vis[lf]=1
                          lf+=1
                          b-=1
                      else:
                          res.append(rf)
                          vis[rf]=1
                          rf-=1
                          a-=1
                  for i in range(1,n+1):
                      if(vis[i]==0):
                          res.append(i)
              print(*res)
                      
                      
                  
  