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.Scanner;
   
  public class MainT {
  	public static Scanner read = new Scanner(System.in);
  	public static void main(String[] args) {
  		int t = read.nextInt();
  		while (t-->0) {
  			solve();
  		}
  	}
   
  	private static boolean solve() {
   
  		int n = read.nextInt();
  		int a = read.nextInt();
  		int b = read.nextInt();
   
  		int left = 1;
  		int right = n;
  		int[] prem = new int[n];
  		int index = 0;
  		if (a+b>n-2 || Math.abs(a-b)>1) {
  			System.out.println(-1);
  			return false;
  		}
  		else {
  			if (a==b) {
  				int count =0;
  				prem[index] = left;
  				left++;
  				index++;
  				while (count<a) {
  				prem[index] = right;
  				right--;
  				index++;
  				prem[index] = left;
  				left++;
  				index++;
  				count++;
  				}
  				for (int i = left; i <= right; i++) {
  					prem[index] = i;
  					index++;
  				}
  			}
  			else if (a>b) {
  				int count= 0;
  				prem[index] = left;
  				index++;
  				left++;
  				while (count<a-1) {
  					prem[index] = right;
  					index++;
  					right--;
  					prem[index] = left;
  					left++;
  					index++;
  					count++;
  				}
  				for (int i = right; i >= left; i--) {
  					prem[index] = i;
  					index++;
  				}
  			}
  			else {
  				prem[index] = right;
  				right--;
  				index++;
  				int count = 0;
  				while (count<b-1) {
  					prem[index] = left;
  					left++;
  					index++;
  					prem[index] = right;
  					right--;
  					index++;
  					count++;
  				}
  				for (int i = left; i <=right ; i++) {
  					prem[index] = i;
  					index++;
   
  				}
  			}
  		}
   
  		for (int i = 0; i < prem.length; i++) {
  			System.out.println(prem[i]);
  		}
  		return true;
  	}
  }
   