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: /*==========================================================================
  *   	AUTHOR:    RonWonWon
  *   	CREATED:   12.12.2021 23:16:04        
  /*==========================================================================*/
  import java.io.*;
  import java.util.*;
  
  public class B {
  	
  	public static void main(String[] args) throws IOException {
  		/*
      in.ini();
      PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
      */
      long startTime = System.nanoTime();
      int t = in.nextInt(), T = 1;  
      while(t-->0) { 
        int n = in.nextInt(), a = in.nextInt(), b = in.nextInt();
        int ar[] = new int[n];
        if(a==0&&b==0){
        	for(int i=1;i<=n;i++) out.print(i+" ");
        	out.println();
        }
       	else if(a-b==1){
       		if(a>(n-1)/2){
       			out.println(-1); continue;
       		}
  				int p = n, q = 1, ind = 0;
  				for(int i=0;i<a;i++){
  					ar[ind] = q; ar[ind+1] = p;
  					ind += 2; p--; q++;
  				}
  				while(ind<n){
  					ar[ind++] = p--;
  				}
       		for(int i : ar) out.print(i+" ");
       		out.println();
       	}
       	else if(b-a==1){
       		if(b>(n-1)/2){
       			out.println(-1); continue;
       		}
  				int p = 1, q = n, ind = 0;
  				for(int i=0;i<b;i++){
  					ar[ind] = q; ar[ind+1] = p;
  					ind += 2; p++; q--;
  				}
  				while(ind<n){
  					ar[ind++] = p++;
  				}
       		for(int i : ar) out.print(i+" ");
       		out.println();
       	}
       	else if(a==b && a<=(n-1)/2){
       		if(a==(n-1)/2&&n%2==1){
       			out.println(-1); continue;
       		}
       		int p=2, q=n, ind = 1;
       		ar[0] = 1;
       		for(int i=0;i<a;i++){
  					ar[ind] = q; ar[ind+1] = p;
  					p++; q--; ind+=2;
       		}
       		while(ind<n) ar[ind++] = p++;
       		for(int i=1;i<n-1;i++){
       			if(ar[i-1]<ar[i]&&ar[i]>ar[i+1]) a--;
       			else if(ar[i-1]>ar[i]&&ar[i]<ar[i+1]) b--;
       		}
       		if(a!=0||b!=0){
       			out.println(-1); continue;
       		}
       		for(int i : ar) out.print(i+" ");
       		out.println();
       	}
       	else{
       		out.println(-1);
       	}
     		//out.println("Case #"+T+": "+ans); T++;
      }
      /*
      startTime = System.nanoTime() - startTime;
      System.err.println("Time: "+(startTime/1000000));
      */
      out.flush();
  	}
  	
  	static FastScanner in = new FastScanner();
    static PrintWriter out = new PrintWriter(System.out);
  	static int oo = Integer.MAX_VALUE;
  	static long ooo = Long.MAX_VALUE;
  	
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer("");
  		
  		void ini() throws FileNotFoundException {
  			br = new BufferedReader(new FileReader("input.txt"));
  		}
  		
  		String next() {
  			while(!st.hasMoreTokens())
  				try { st = new StringTokenizer(br.readLine()); }
  				catch(IOException e) {}
  			return st.nextToken();
  		}
  		
  		String nextLine(){
  			try{ return br.readLine(); } 
  			catch(IOException e) { } return "";
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		int[] readArray(int n) {
  			int a[] = new int[n];
  			for(int i=0;i<n;i++) a[i] = nextInt();
  			return a;
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		
  		long[] readArrayL(int n) {
  			long a[] = new long[n];
  			for(int i=0;i<n;i++) a[i] = nextLong();
  			return a;
  		}
  		
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		
  		double[] readArrayD(int n) {
  			double a[] = new double[n];
  			for(int i=0;i<n;i++) a[i] = nextDouble();
  			return a;
  		}
  	}
  
  	static final Random random = new Random();
  
  	static void ruffleSort(int[] a){
  		int n = a.length;
  		for(int i=0;i<n;i++){
  			int j = random.nextInt(n), temp = a[j];
  			a[j] = a[i]; a[i] = temp;
  		}
  		Arrays.sort(a); 	
  	}
  	
  	static void ruffleSortL(long[] a){
  		int n = a.length;
  		for(int i=0;i<n;i++){
  			int j = random.nextInt(n);
  			long temp = a[j];
  			a[j] = a[i]; a[i] = temp;
  		}
  		Arrays.sort(a); 	
  	}
  }
  