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.BufferedReader;
  import java.io.InputStreamReader;
  import java.util.*;
  
  public class Main {
      public static void main(String[] args) throws Exception {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  
          StringTokenizer st = new StringTokenizer(br.readLine());
          int t = Integer.parseInt(st.nextToken());
  
          StringBuilder sb = new StringBuilder();
  
          while (t != 0) {
              st = new StringTokenizer(br.readLine());
              int n = Integer.parseInt(st.nextToken());
              int a = Integer.parseInt(st.nextToken());
              int b = Integer.parseInt(st.nextToken());
  
              if(a+b+2>n || Math.abs(a-b)>1){
                  sb.append("-1").append("\n");
              }
              else{
                  boolean reverse= a < b;
                  sb.append(getPermutation(n,Math.max(a,b),Math.min(a,b),reverse)).append("\n");
              }
              t--;
          }
          System.out.println(sb);
      }
  
      private static String getPermutation(int n, int a, int b, boolean reverse) {
          int total=a+b+2;
          int start=n;
          int end=n-total+1;
          int remain=end-1;
  
          StringBuilder sb=new StringBuilder();
  
          boolean toggle=true;
          for(int i=0;i<total;i++){
              if(toggle){
                  sb.append(end).append(" ");
                  end++;
                  toggle= false;
              }
              else{
                  sb.append(start).append(" ");
                  start--;
                  toggle= true;
              }
          }
          StringBuilder sb2 = new StringBuilder();
          for(int i=1;i<=remain;i++){
              sb2.append(i).append(" ");
          }
  
          sb2.append(sb.toString().trim());
          if(reverse){
              return getReverse(n,sb2.toString());
          }
          return sb2.toString();
      }
  
      public static String getReverse(int n,String s){
          StringBuilder sb=new StringBuilder();
          for(String c:s.split(" ")){
              sb.append(n-Integer.parseInt(c)+1).append(" ");
          }
          return sb.toString().trim();
      }
  
  
      public static void swap(int[] ar,int i,int j){
          int temp=ar[i];
          ar[i]=ar[j];
          ar[j]=temp;
      }
  
      public static class Pair{
          int x;
          int y;
  
          Pair(int x,int y){
              this.x=x;
              this.y=y;
          }
  
          public int getX(){
              return x;
          }
  
          public int getY(){
              return y;
          }
      }
  }
  
  