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;
  int main(void) {
    int test_case;
    cin >> test_case;
    for (int t = 0; t < test_case; t++) {
      int n, a, b;
      vector<int> arr;
      cin >> n >> a >> b;
      if ((a + b + 2 > n) || (abs(a - b) > 1)) {
        cout << -1 << endl;
        continue;
      }
      for (int i = 1; i <= n; i++) arr.push_back(i);
      if (b > a) {
        int idx = 2;
        for (int i = 0; i < a; i++) {
          swap(arr[idx], arr[idx + 1]);
          idx += 2;
        }
        swap(arr[0], arr[1]);
      } else if (b == a) {
        int idx = 1;
        for (int i = 0; i < a; i++) {
          swap(arr[idx], arr[idx + 1]);
          idx += 2;
        }
      } else {
        int idx = 1;
        for (int i = 0; i < b; i++) {
          swap(arr[idx], arr[idx + 1]);
          idx += 2;
        }
        swap(arr[n - 2], arr[n - 1]);
      }
      for (int i : arr) cout << i << " ";
      cout << endl;
    }
  }
  