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
  import os.path
   
  if(os.path.exists('input.txt')) :
      sys.stdin = open("input.txt", "r")
      sys.stdout = open("output.txt", "w")
      sys.stderr = open("error.txt", "w")
      
  depth = 200001
  mod = 1000000007 
  lim = mod * mod
  sys.setrecursionlimit(depth) 
   
  linp = lambda: list(minp())
  minp = lambda: map(int, input().split())
   
  from math import inf, ceil, sqrt, log2, gcd
  from collections import defaultdict, deque
  
  dxy = [(1, 0),(-1, 0),(0, 1),(0, -1)]
  
  def expo(a, b) :
      if not b : return 1
      v = expo(a, b//2)
      v = (v * v) % mod
      if b%2 : v = (v * a) % mod
      return v
  
  def modinv(a, b) :
      d = expo(b, mod-2)
      v = (a * d) % mod
      return v
  
  def checkPrime(n) :
      if n < 4 : return True
      if n%2 == 0 : return False
      for i in range(3, int(sqrt(n))+1, 2) :
          if n%i == 0 :
              return False
      return True    
  
  def highestPowerof2(x):
      x |= x >> 1
      x |= x >> 2
      x |= x >> 4
      x |= x >> 8
      x |= x >> 16
  
      return x ^ (x >> 1)
  
  for _ in range(int(input())) :
      n, a, b = minp()
  
      if a+b+2 > n or abs(a-b) > 1 :
          print(-1)
          continue 
  
      if a >= b : v = [i+1 for i in range(n)]
      else : v = [n-i for i in range(n)]
  
  
      for i in range(1, n-2, 2) :
          if not a or not b : break
          
          v[i], v[i+1] = v[i+1], v[i]
          a, b = a - 1, b - 1
  
      if a or b :
          v[-1], v[-2] = v[-2], v[-1]
  
      print(*v)
  
  
  
  