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: //package Div2.B;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  
  public class BuildThePermutation {
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          int t=Integer.parseInt(br.readLine());
          while(t>0){
              String []nab=br.readLine().split(" ");
              int n=Integer.parseInt(nab[0]);
              int a=Integer.parseInt(nab[1]);
              int b=Integer.parseInt(nab[2]);
              if(a+b>n-2 || abs(a-b)>1)
                  System.out.println(-1);
              else{
                  StringBuilder ans=new StringBuilder();
                  int min=1;
                  int max=n;
                  if(a>=b){
                      ans.append(min).append(" ");
                      min++;
                      for(int i=1;i<=a+b;i++){
                          if(i%2!=0) {
                              ans.append(max).append(" ");
                              max--;
                          }else{
                              ans.append(min).append(" ");
                              min++;
                          }
                      }
  
                  }else{
                      ans.append(max).append(" ");
                      max--;
                      for(int i=1;i<=a+b;i++){
                          if(i%2!=0) {
                              ans.append(min).append(" ");
                              min++;
                          }
                          else {
                              ans.append(max).append(" ");
                              max--;
                          }
                      }
  
                  }
  
  
                  if(a>b){
                      while(max>=min){
                          ans.append(max).append(" ");
                          max--;
                      }
                  }else{
                      while(min<=max){
                          ans.append(min).append(" ");
                          min++;
                      }
                  }
                  System.out.println(ans);
              }
              t--;
          }
      }
      private static int abs(int x){
          return x<0?-x:x;
      }
  }
  