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.*;
  public class Main {
  	public static void main(String[] args){
  		Scanner sc = new Scanner(System.in);
  
  		// long mod = 1_000_000_007L;
  		// long mod = 998_244_353L;
  
  		int t = sc.nextInt();
  
  		for ( int zzz=0; zzz<t; zzz++ ) {
  
  			int n = sc.nextInt();
  			int a = sc.nextInt();
  			int b = sc.nextInt();
  			if ( Math.max(a,b)-Math.min(a,b)>1 ) {
  				System.out.println(-1);
  				continue;
  			}
  			if ( a+b>n-2 ) {
  				System.out.println(-1);
  				continue;
  			}
  
  			Deque<String> q = new ArrayDeque<>();
  			for ( int i=1; i<=n; i++ ) {
  				q.addLast(String.valueOf(i));
  			}
  
  			String[] s = new String[n];
  
  			if ( a==0 && b==0 ) {
  				for ( int i=0; i<n; i++ ) {
  					s[i] = q.removeFirst();
  				}
  				System.out.println(String.join(" ", s));
  				continue;
  			}
  			for ( int i=1; i<=Math.max(a,b)-1; i++ ) {
  				if ( a>=b ) {
  					s[i*2-1] = q.removeLast();
  					s[i*2] = q.removeFirst();
  				} else {
  					s[i*2-1] = q.removeFirst();
  					s[i*2] = q.removeLast();
  				}
  			}
  
  			for ( int i=Math.max(a,b)*2-1; i<n; i++ ) {
  				if ( a>b ) {
  					s[i] = q.removeLast();
  				} else if ( a<b ) {
  					s[i] = q.removeFirst();
  				} else {
  					if ( i==a*2-1 ) {
  						s[i] = q.removeLast();
  					} else {
  						s[i] = q.removeFirst();
  					}
  				}
  			}
  			s[0] = q.removeFirst();
  
  			System.out.println(String.join(" ", s));
  		}
  	}
  }
  