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 java.io.*;
  import java.util.*;
  
  public class BMain {
  
  	public static void main(String[] args) {
  		QuickReader in = new QuickReader(System.in);
  		
  		try(PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));)
  		{
  			new BMain().solve(in, out);
  		}
  	}
  	
  	public void solve(QuickReader in, PrintWriter out)
  	{
  		int T = in.nextInt();
  		while(T-->0)
  		{
  			int n = in.nextInt();
  			int a = in.nextInt();
  			int b = in.nextInt();
  			
  			ArrayList<Integer> res = new ArrayList<Integer>();
  			
  			if( a+b == 0 )
  			{
  				for(int i=0;i<n;i++)
  					res.add(i+1);
  			}
  			else if( Math.abs(a-b)<=1 && n>=2+a+b )
  			{
  				for(int i=0;i<n;i++)
  					res.add(0);
  				
  				int sz = a+b+2;
  				if(a>=b)
  				{
  					int lo = 1;
  					int hi = sz;
  					for(int i=1;i<sz;i+=2)
  						res.set(i, hi--);
  					for(int i=0;i<sz;i+=2)
  						res.set(i, lo++);
  				}
  				else
  				{
  					int lo = 1;
  					int hi = sz;
  					for(int i=1;i<sz;i+=2)
  						res.set(i, lo++);
  					for(int i=0;i<sz;i+=2)
  						res.set(i, hi--);
  				}
  				
  				if(res.get(sz-2) < res.get(sz-1))
  				{
  					for(int i=sz;i<n;i++)
  						res.set(i, i+1);
  				}
  				else
  				{
  					int inc = n-sz;
  					for(int i=0;i<sz;i++)
  						res.set(i, res.get(i) + inc );
  					for(int i=sz;i<n;i++)
  						res.set(i, inc+(sz-i));
  				}
  			}
  			else
  				res.add(-1);
  			
  			for(int i: res)
  			{
  				out.print(i);
  				out.print(" ");
  			}
  			out.println();
  			out.flush();
  			
  		}
  		
  	}
  }
  
  class QuickReader
  {
  	BufferedReader in;
  	StringTokenizer token;
  	
  	public QuickReader(InputStream ins)
  	{
  		in=new BufferedReader(new InputStreamReader(ins));
  		token=new StringTokenizer("");
  	}
  
  	public boolean hasNext()
  	{
  		while (!token.hasMoreTokens())
  		{
  			try
  			{
  				String s = in.readLine();
  				if (s == null) return false;
  				token = new StringTokenizer(s);
  			} catch (IOException e)
  			{
  				throw new InputMismatchException();
  			}
  		}
  		return true;
  	}
  
  	public String next()
  	{
  		hasNext();
  		return token.nextToken();
  	}
  
  	public int nextInt()
  	{
  		return Integer.parseInt(next());
  	}
  
  	public int[] nextInts(int n)
  	{
  		int[] res = new int[n];
  		for (int i = 0; i < n; i++)
  			res[i] = nextInt();
  		return res;
  	}
  
  	public long nextLong() {
  		return Long.parseLong(next());
  	}
  	
  	public long[] nextLongs(int n)
  	{
  		long[] res = new long[n];
  		for (int i = 0; i < n; i++)
  			res[i] = nextLong();
  		return res;
  	}
  	
  }
  
  