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.util.*;
  import java.io.*;
  
  public class Solution {
  	public static void reverse(int arr[]) {
  		int l=0,r=arr.length-1;
  		while(l<r) {
  			int t=arr[l];
  			arr[l++]=arr[r];
  			arr[r--]=t;
  		}
  	}
  	public static void main(String args[]) {
  	   Scanner sc=new Scanner(System.in);
  	   int t=sc.nextInt();
  	  
  	   
  	   while(t-->0) {
  		   int n=sc.nextInt();
  		   int a=sc.nextInt();
  		   int b=sc.nextInt();
  		   int arr[]=new int[n];
  		   for(int i=1;i<=n;i++) arr[i-1]=i;
  		   
  		  if(Math.abs(a-b)>1 || a+b>n-2) {
  			  System.out.println(-1);
  		  }
  		  else {
  			  if(a==b) {
  				  for(int i=1;a-->0;i+=2) {
  					  int temp=arr[i];
  					  arr[i]=arr[i+1];
  					  arr[i+1]=temp;
  				  }
  			  }
  			  else if(a>b) {
  				  for(int i=(n-(a*2));i<n-1;i+=2) {
  					  int temp=arr[i];
  					  arr[i]=arr[i+1];
  					  arr[i+1]=temp;
  				  }
  				  }
  			  else {
  				  reverse(arr);
  				  for(int i=(n-(b*2))+1;i<n;i+=2) {
  					  int temp=arr[i];
  					  arr[i]=arr[i-1];
  					  arr[i-1]=temp;
  				  }
  			  }
  			  for(int x:arr) System.out.print(x+" ");
  			  System.out.println();
  		  }
  		  
  		   
  		   
  	  }
  	   
  }
  	}