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: #include <bits/stdc++.h>
  const double PI = 3.1415926535897932384626433;
  const int KL = 1e6;
  const long long MOD = 1e9 + 7;
  using namespace std;
  int q, x, y, z, n, m, a, b, ans;
  string t;
  vector<int> vec;
  int main() {
    scanf("%d", &q);
    while (q--) {
      scanf("%d", &n);
      scanf("%d", &a);
      scanf("%d", &b);
      if (a + b <= n - 2 && abs(a - b) <= 1) {
        if (a == b + 1) {
          int cnt = n - 1;
          while (a > 0) {
            printf("%d %d ", cnt, cnt + 1);
            cnt -= 2;
            a -= 1;
          }
          for (int i = cnt + 1; i >= 1; i--) printf("%d ", i);
          printf("\n");
        } else if (a == b) {
          int cnt = a + b;
          while (a > 0) {
            printf("%d %d ", cnt, cnt + 1);
            cnt -= 2;
            a--;
          }
          printf("1 ");
          for (int i = b + b + 2; i <= n; i++) printf("%d ", i);
          printf("\n");
        } else {
          int cnt = 2;
          while (b > 0) {
            printf("%d %d ", cnt, cnt - 1);
            cnt += 2;
            b--;
          }
          for (int i = cnt - 1; i <= n; i++) printf("%d ", i);
          printf("\n");
        }
      } else
        printf("-1\n");
    }
    return 0;
  }
  