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>
  using namespace std;
  const int inf = 0x3f3f3f3f;
  const double eps = 1e-8;
  const double PI = acos(-1);
  int T;
  int n, a, b;
  bool check() {
    if (a + b > n - 2) return false;
    if (abs(a - b) > 1) return false;
    int t = ceil(1.0 * (n - 2) / 2);
    if (a > t || b > t) return false;
    return true;
  }
  void print() {
    if (a > b) {
      int i = 1, j = n;
      int t = a;
      while (t--) printf("%d %d ", i++, j--);
      while (i <= j) printf("%d ", j--);
      puts("");
    } else if (b > a) {
      int i = 1, j = n;
      int t = b;
      while (t--) printf("%d %d ", j--, i++);
      while (i <= j) printf("%d ", i++);
      puts("");
    } else {
      int i = 1, j = n;
      int t = a;
      while (t--) printf("%d %d ", i++, j--);
      while (i <= j) printf("%d ", i++);
      puts("");
    }
  }
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d%d%d", &n, &a, &b);
      if (!check()) {
        puts("-1");
        continue;
      }
      print();
    }
    return 0;
  }
  