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;
  void solve() {
    long long n, a, b;
    cin >> n >> a >> b;
    vector<long long> res;
    for (long long i = 1; i <= n; i++) res.push_back(i);
    if ((abs(a - b)) > 1 || ((a + b) > n - 2)) {
      cout << -1 << endl;
      return;
    } else {
      long long i = 1, lim = n;
      if (a > b) {
        res[n - 1] = res[n - 2];
        res[n - 2] = n;
        a--;
        lim = n - 2;
        i = 1;
      }
      if (b > a) {
        res[0] = 2;
        res[1] = 1;
        b--;
        i = 2;
      }
      for (; i < lim && a > 0 && b > 0; i += 2) {
        res[i + 1] = res[i];
        res[i] = i + 2;
        a--, b--;
      }
      for (long long i = 0; i < n; i++) cout << res[i] << " ";
      cout << endl;
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  