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 long long MOD = 1e9 + 7;
  const long long infi = INT_MAX;
  const long long infll = LLONG_MAX;
  const long double PI = 3.1415926535897932384626;
  void solve(long long test_case) {
    long long n, a, b;
    cin >> n >> a >> b;
    if (a + b + 2 > n) {
      cout << -1 << "\n";
      return;
    }
    if (abs(a - b) > 1) {
      cout << -1 << "\n";
      return;
    }
    vector<long long> arr(n);
    for (long long i = 0; i < n; i++) arr[i] = i + 1;
    if (a > b) {
      for (long long i = n - 1; i - 1 >= 0; i -= 2) {
        swap(arr[i], arr[i - 1]);
        if (--a == 0) break;
      }
    } else if (b > a) {
      for (long long i = 0; i + 1 < n; i += 2) {
        swap(arr[i], arr[i + 1]);
        if (--b == 0) break;
      }
    } else {
      for (long long i = 1; i + 1 < n; i += 2) {
        if (a == 0) break;
        swap(arr[i], arr[i + 1]);
        a--;
      }
    }
    for (auto x : arr) cout << x << " ";
    cout << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ios_base ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long tc = 1;
    cin >> tc;
    while (tc--) solve(tc);
    return 0;
  }
  