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.time.Year;
  import java.util.Scanner;
  import java.util.*;
  public class test2 {
  
  
      public static void main(String[] args) {
  
          Scanner in = new Scanner(System.in);
  //            测试次数
          int testnum = in.nextInt();
          for (int i = 0; i< testnum; i++) {
              int n = in.nextInt();
              int b = in.nextInt();
              int a = in.nextInt();
              int na = a;
              int nb = b;
              int[] ans = new int[n];
              if(a>(n-1)/2||b>(n-1)/2||a+b>n-2||Math.abs(a-b)>1){
                  System.out.println(-1);
                  continue;
              }
              int min = 1;
              int max = n;
              if(a>b){
                  int start = 1;
                  while ((a-->0)){
                      ans[start] = min++;
                      ans[start - 1] = max--;
                      start += 2;
                  }
                  if(na==nb){
                      for (int j = start-1; j <n ; j++) {
                          ans[j] = max--;
                      }
                  }else{
                      for (int j = n-1; j >=start-1 ; j--) {
                          ans[j] = max--;
                      }
                  }
              }else{
                  int start = 1;
                  while ((b-->0)){
                      ans[start] = max--;
                      ans[start - 1] = min++;
                      start += 2;
                  }
                  if(na==nb){
                      for (int j = start-1; j <n ; j++) {
                          ans[j] = min++;
                      }
                  }else{
                      for (int j = n-1; j >=start-1 ; j--) {
                          ans[j] = min++;
                      }
                  }
              }
              for (int j = 0; j < n; j++) {
                  System.out.print(ans[j]+" ");
              }
              System.out.println();
          }
  
  
      }
  
  
  
  }
  