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 rough {
       public static void main(String[] args) {
           Scanner sc = new Scanner(System.in);
           int t = sc.nextInt();
           for(int tt = 0; tt < t; tt++){
              int n = sc.nextInt(), a = sc.nextInt(), b = sc.nextInt();
               int[] ans = new int[n];
               if(a+b > n-2){
                  System.out.println("-1");
                  continue;
               } else if(Math.abs(a - b ) > 1){
                  System.out.println("-1");
                  continue;
               }
               if(a > b){
                  int curr = n;
                  for(int i = 1, cnt = 0; cnt < a; cnt++,i+=2){
                      ans[i] = curr; curr--;
                  }
                  int right = curr;
                  curr = 1;
                  for(int i = 2,cnt = 0; cnt < b; cnt++,i+=2){
                      ans[i] = curr; curr++;
                  }
                  
                  ans[0] = right--;
                  for(int i = a+b+1; i < n; i++){
                      ans[i] = right--;
                  }
               }
  
               if(b > a){
                  int curr = 1;
                  for(int i = 1,cnt = 0; cnt < b; i+=2, cnt++){
                      ans[i] = curr; curr++;
                  }
  
                  int right = curr;
                  curr = n;
                  for(int i = 2,cnt = 0; cnt < a; i+=2,cnt++){
                      ans[i] = curr; curr--;
                  }
  
                  
                  ans[0] = right++; 
                  for(int i = a+b+1; i < n; i++){
                      ans[i] = right++;
                  }
               }
  
               if(a == b){
                  int curr = n;
                  for(int i = 1, cnt = 0; cnt < a; cnt++,i+=2){
                      ans[i] = curr; curr--;
                  }
                  curr = 1;
                  for(int i = 2,cnt = 0; cnt < b; cnt++,i+=2){
                      ans[i] = curr; curr++;
                  }
                  int left = curr;
                  ans[0] = left++;
                  for(int i = a+b+1; i < n; i++){
                      ans[i] = left++;
                  }
               }
  
               for(int i = 0; i < n; i++){
                  System.out.println(ans[i]);
               }
  
           }
       }
   }