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;
  constexpr int N = 3e5 + 10;
  int arr[N];
  void solve() {
    int n, a, b;
    cin >> n >> a >> b;
    if (a + b + 2 > n) {
      cout << -1 << endl;
      return;
    }
    if (abs(a - b) > 1) {
      cout << -1 << endl;
      return;
    }
    if (a == b) {
      for (int i = a + b + 2; i < n; i++) {
        arr[i] = i + 1;
      }
      for (int i = 0; i <= b; i++) {
        arr[i * 2 + 1] = a + i + 2;
      }
      for (int i = 0; i <= a; i++) {
        arr[i * 2] = i + 1;
      }
    } else if (a < b) {
      for (int i = 0; i <= b; i++) {
        arr[i * 2] = b + i + 1;
      }
      for (int i = 0; i <= a; i++) {
        arr[i * 2 + 1] = i + 1;
      }
      for (int i = a + b + 2; i < n; i++) {
        arr[i] = i + 1;
      }
    } else {
      int cnt = n - (a + b + 2);
      for (int i = 0; i <= a; i++) {
        arr[i * 2] = cnt + i + 1;
      }
      for (int i = 0; i <= b; i++) {
        arr[i * 2 + 1] = n - i;
      }
      for (int i = n - cnt; i < n; i++) {
        arr[i] = n - i;
      }
    }
    if (a == 0 && b == 0) {
      for (int i = 0; i < n; i++) {
        arr[i] = i + 1;
      }
    }
    for (int i = 0; i < n; i++) {
      cout << arr[i] << ' ';
    }
    cout << endl;
  }
  int main() {
    int T;
    cin >> T;
    while (T--) solve();
    return 0;
  }
  