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 sys
  from math import factorial, gcd
  #from math import comb, perm 	
  from collections import Counter, deque, defaultdict
  from bisect import bisect_left, bisect_right 
  from heapq import heappop, heappush, heapify, nlargest, nsmallest
  from itertools import groupby
  from copy import deepcopy
  MOD = 10**9+7
  INF = float('inf')
  
  # import io,os
  # input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
  # input = sys.stdin.buffer.readline
  input = sys.stdin.readline
  rl = lambda : list(map(int, input().split()))
  rs = lambda : sys.stdin.readline().strip()
  
  def ceil(a, b):
  	return (a + b - 1) // b
  
  for _ in range(int(input())):
  	n, a, b = rl()
  	if ceil(n, 2) <= a or ceil(n, 2) <= b or n - 2 < a + b or abs(a-b) >= 2:
  		print(-1)
  		continue
  
  	res = [i for i in range(1, n+1)]
  	if a > b:
  		i = n - 1
  		while a:
  			res[i], res[i-1] = res[i-1], res[i]
  			a -= 1
  			i -= 2
  	elif a < b:
  		res = res[::-1]
  		i = n - 1
  		while b:
  			res[i], res[i-1] = res[i-1], res[i]
  			b -= 1
  			i -= 2
  	else:
  		i = 1
  		while a:
  			res[i], res[i+1] = res[i+1], res[i]
  			i += 2
  			a -= 1
  
  	print(*res)