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 (a+b)>(n-2):
  		print(-1)
  	else:
  		if abs(a-b)>1:
  			print(-1)
  		else:
  			li=[]
  			ma=n
  			mi=1
  			if a>b:
  				maflag=1
  				for i in range(a+b):
  					if maflag==1:
  						li.append(ma)
  						ma-=1
  						maflag=0
  					else:
  						li.append(mi)
  						mi+=1
  						maflag=1
  				li.insert(0,ma)
  				ma-=1
  				for i in range(n-a-b-1):
  					li.append(ma)
  					ma-=1
  			elif a<b:
  				miflag=1
  				for i in range(a+b):
  					if miflag==1:
  						li.append(mi)
  						mi+=1
  						miflag=0
  					else:
  						li.append(ma)
  						ma-=1
  						miflag=1
  				li.insert(0,mi)
  				mi+=1
  				for i in range(n-a-b-1):
  					li.append(mi)
  					mi+=1
  			elif a==b and a!=0:
  				maflag=1
  				for i in range(a+b):
  					if maflag==1:
  						li.append(ma)
  						ma-=1
  						maflag=0
  					else:
  						li.append(mi)
  						mi+=1
  						maflag=1
  				li.insert(0,ma)
  				for i in range(n-a-b-1):
  					li.append(mi)
  					mi+=1
  			else:
  				for i in range(1,n+1):
  					li.append(i)
  			print(*li)
  			