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: t=int(input())
  for i in range(t):
      n,a,b=map(int,input().split())
      if abs(a-b)>1 or (a+b)>(n-2):
          print(-1)
      else:
          if a>b:
              lst = [0] * ( (a + b) + 1)
              lsts = []
              lstr = []
              for y in range(1, n + 1):
                  lsts.append(y)
              for l in range(n, 0, -1):
                  lstr.append(l)
              c=0
              d=1
              while c!=a:
                  lst[d]=lstr[c]
                  d+=2
                  c+=1
              d=2
              c=0
              while c!=b:
                  lst[d]=lsts[c]
                  d+=2
                  c+=1
              lsto=[]
              for x in range(b+1,n-a+1):
                  lsto.append(x)
              lsto.sort(reverse=True)
              lst[0]=lsto[0]
              lst=lst+lsto[1:]
              print(*lst)
          elif b>a:
              lst = [0] * ((a + b) + 1)
              lsts = []
              lstr = []
              for y in range(1, n + 1):
                  lsts.append(y)
              for l in range(n, 0, -1):
                  lstr.append(l)
              c = 0
              d = 1
              while c != b:
                  lst[d] = lsts[c]
                  d += 2
                  c += 1
              d = 2
              c = 0
              while c !=a:
                  lst[d] = lstr[c]
                  d += 2
                  c += 1
              lsto = []
              for x in range(b + 1, n - a + 1):
                  lsto.append(x)
              lsto.sort()
              lst[0] = lsto[0]
              lst = lst + lsto[1:]
              print(*lst)
          else:
              lst = [0] * ((a + b) + 1)
              lsts = []
              lstr = []
              for y in range(1, n + 1):
                  lsts.append(y)
              for l in range(n, 0, -1):
                  lstr.append(l)
  
              c = 0
              d = 1
              while c != b:
                  lst[d] = lsts[c]
                  d += 2
                  c += 1
              d = 2
              c = 0
              while c != a:
                  lst[d] = lstr[c]
                  d += 2
                  c += 1
              lsto = []
              for x in range(b + 1, n - a + 1):
                  lsto.append(x)
              lsto.sort(reverse=True)
              lst[0] = lsto[0]
              lst = lst + lsto[1:]
              print(*lst)