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.*;
      import java.io.*;
      
      public class MyCpClass{
          public static void main(String []args) throws IOException{
              BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
              StringBuilder sb = new StringBuilder();
              
              int T = Integer.parseInt(br.readLine().trim());
              while(T-- > 0){
                  String []str = br.readLine().trim().split(" ");
                  int n = Integer.parseInt(str[0]);
                  int a = Integer.parseInt(str[1]);
                  int b = Integer.parseInt(str[2]);
                  
                  int []res = new int[n];
                  if(n % 2 == 0 && Math.max(a, b) > (n+1)/2-1){
                      sb.append("-1\n");
                      continue;
                  }
                  if(n % 2 != 0 && Math.max(a, b) > (n+1)/2-1){
                      sb.append("-1\n");
                      continue;
                  }
                  if(n % 2 != 0 && a == (n+1)/2-1 && b == (n+1)/2-1){
                      sb.append("-1\n");
                      continue;
                  }
                  if(Math.abs(a-b) > 1){
                      sb.append("-1\n");
                      continue;
                  }
                  if(a > b){
                      int x = 1;
                      for(int i=0; i<2*a; i+=2)
                          res[i] = x++;
                      for(int i=n-1; i>=2*a; i--)
                          res[i] = x++;
                      for(int i=1; i<2*a; i+=2)
                          res[i] = x++;
                      
                  }
                  else if(a < b){
                      int x = 1;
                      for(int i=1; i<2*b; i+=2)
                          res[i] = x++;
                      for(int i=2*b; i<n; i++)
                          res[i] = x++;
                      for(int i=0; i<2*b; i+=2)
                          res[i] = x++;
                  }
                  else{
                      int x = 1;
                      for(int i=0; i<=2*a; i+=2)
                          res[i] = x++;
                      for(int i=1; i<=2*a; i+=2)
                          res[i] = x++;
                      for(int i=2*a+1; i<n; i++)
                          res[i] = x++;
                  }
                  
                  for(int i=0; i<n; i++)
                      sb.append(res[i] + " ");
                  sb.append("\n");
              }
              System.out.println(sb);
          }
      }