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 N = 1000005;
  int t, n, a, b, x[N];
  int main() {
    scanf("%d", &t);
    while (t--) {
      scanf("%d%d%d", &n, &a, &b);
      for (int i = 0; i <= n; ++i) x[i] = 0;
      if (abs(a - b) > 1 || n < a + b + 2) {
        printf("-1\n");
        continue;
      } else {
        if (a > b) {
          int tmp = n - (a + b + 2);
          for (int i = 1; i <= tmp; ++i) x[i] = i;
          for (int i = tmp + 1; i <= n; i += 2) x[i] = ((i - tmp) / 2 + 1) + tmp;
          for (int i = tmp + 2; i <= n; i += 2)
            x[i] = (n - tmp) - (x[i - 1] - tmp) + 1 + tmp;
        } else {
          int tmp = n - (a + b + 2);
          for (int i = 2 + tmp; i <= n; i += 2) x[i] = (i - tmp) / 2;
          for (int i = 1 + tmp; i <= n; i += 2) x[i] = (n - tmp) - x[i - 1];
          for (int i = 1; i <= tmp; ++i) x[i] = 2 * n - tmp + 1 - (i + n - tmp);
        }
      }
      for (int i = 1; i <= n; ++i) printf("%d ", x[i]);
      printf("\n");
    }
    return 0;
  }
  