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.*;
  import static java.lang.Math.*;
  
  public class B{
  	
  	static final int up = -1, down = 1;
  	
  	public static void main(String[] args) throws IOException{
  		// br = new BufferedReader(new FileReader(".in"));
  		// out = new PrintWriter(new FileWriter(".out"));
  		// new Thread(null, new (), "fisa balls", 1<<28).start();
  		
  		int t =readInt();
  		while(t-->0) {
  			int n =readInt();
  			int a =readInt(), b=readInt();
  			a^=b;
  			b^=a;
  			a^=b;
  			if (a+b > n-2 || abs(a-b) > 1) {
  				out.println(-1);
  				continue;
  			}
  			int lp = 1, rp = a+b;
  			// a is min in my code
  			ArrayDeque<Integer> q = new ArrayDeque<Integer>();
  			if (a == 0 && b == 0) {
  				for (int i = 1; i <= n; i++) out.print(i + " ");
  				out.println();
  			}
  			else if (a>b) {
  				for (int i = 0; i < a+b; i++) {
  					if (i%2==0)q.add(lp++);
  					else q.add(rp--);
  				}
  				for (int j = a+b+1; j <= n; j++) {
  					if (j%2==0) q.addFirst(j);
  					else q.addLast(j);
  				}
  				for (int x: q) out.print(x + " ");
  				out.println();
  			}
  			else if (a<b) {
  				for (int i = 0; i < a+b; i++) {
  					if (i%2==1)q.add(lp++);
  					else q.add(rp--);
  				}
  				int min = 0;
  				while(q.size() < n) {
  					if (min%2==0) q.addFirst(min--);
  					else q.addLast(min--);
  				}
  				for (int x: q) out.print(x-min + " ");
  				out.println();
  			}
  			else {
  				lp = 1;
  				rp = n;
  				for (int i = 0; i < a+b; i++) {
  					if (i%2==0)q.add(lp++);
  					else q.add(rp--);
  				}
  				for (int i = a+b; i < n; i++) {
  					if (i%2==0)q.addFirst(lp++);
  					else q.addLast(rp--);
  				}
  				for (int x: q) out.print(x + " ");
  				out.println();
  			}		
  			
  		}
  		
  		out.close();
  	}
  	
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  	static StringTokenizer st = new StringTokenizer("");
  	static String read() throws IOException{
  		while (!st.hasMoreElements()) st = new StringTokenizer(br.readLine());
  		return st.nextToken();
  	}
  	public static int readInt() throws IOException{return Integer.parseInt(read());}
  	public static long readLong() throws IOException{return Long.parseLong(read());}
  }