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>
  #pragma optimize("SEX_ON_THE_BEACH")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC optimize("unroll-all-loops")
  #pragma GCC optimize("O3")
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("fast-math")
  #pragma GCC optimize("no-stack-protector")
  using ll = long long int;
  using ull = unsigned long long int;
  using dd = double;
  using ldd = long double;
  namespace someUsefull {
  template <typename T1, typename T2>
  inline void checkMin(T1& a, T2 b) {
    if (a > b) a = b;
  }
  template <typename T1, typename T2>
  inline void checkMax(T1& a, T2 b) {
    if (a < b) a = b;
  }
  const int _bitfunctions_size_ = 16;
  const int _bitfunctions_mask_ = (1 << _bitfunctions_size_) - 1;
  char* _bits_count_;
  inline void _build_bits_count_() {
    _bits_count_ = new char[1 << _bitfunctions_size_];
    for (int i = 1; i < (1 << _bitfunctions_size_); ++i) {
      _bits_count_[i] = _bits_count_[i >> 1];
      if (i & 1) ++_bits_count_[i];
    }
  }
  template <typename T1 = char, typename T2 = int>
  inline T1 popcnt(T2 x) {
    T1 ans = 0;
    while (x) {
      ans += _bits_count_[x & _bitfunctions_mask_];
      x >>= _bitfunctions_size_;
    }
    return ans;
  }
  }  // namespace someUsefull
  namespace operators {
  template <typename T1, typename T2>
  std::istream& operator>>(std::istream& in, std::pair<T1, T2>& x) {
    in >> x.first >> x.second;
    return in;
  }
  template <typename T1, typename T2>
  std::ostream& operator<<(std::ostream& out, std::pair<T1, T2> x) {
    out << x.first << " " << x.second;
    return out;
  }
  template <typename T1>
  std::istream& operator>>(std::istream& in, std::vector<T1>& x) {
    for (auto& i : x) in >> i;
    return in;
  }
  template <typename T1>
  std::ostream& operator<<(std::ostream& out, std::vector<T1>& x) {
    for (auto& i : x) out << i << " ";
    return out;
  }
  }  // namespace operators
  using namespace std;
  using namespace operators;
  using namespace someUsefull;
  void solve(int test = 0) {
    int n, a, b;
    cin >> n >> a >> b;
    bool sw = 0;
    if (a < b) {
      swap(a, b);
      sw = 1;
    }
    if (abs(a - b) > 1) {
      cout << -1 << '\n';
      return;
    }
    if (a > (n - 1) / 2) {
      cout << -1 << '\n';
      return;
    }
    if (b > (n - 2) / 2) {
      cout << -1 << '\n';
      return;
    }
    vector<int> arr(n);
    arr[0] = 1;
    set<int> have;
    for (int i = 1; i < n; ++i) {
      have.insert(i + 1);
    }
    if (a == 0) {
      for (int i = 0; i < n; ++i) {
        cout << i + 1 << " ";
      }
      cout << '\n';
      return;
    }
    for (int i = 1; i < n; ++i) {
      if (i % 2) {
        arr[i] = *have.rbegin();
        have.erase(arr[i]);
        a--;
      } else {
        arr[i] = *have.begin();
        have.erase(arr[i]);
        b--;
      }
      if (a == 0 && b == 0) {
        bool fl = i % 2;
        i++;
        for (; i < n; ++i) {
          if (fl)
            arr[i] = *have.rbegin();
          else
            arr[i] = *have.begin();
          have.erase(arr[i]);
        }
      }
    }
    if (sw) {
      for (int i = 0; i < n; ++i) arr[i] = n + 1 - arr[i];
    }
    cout << arr << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);
    _build_bits_count_();
    int t = 1;
    cin >> t;
    for (int i = 0; i < t; ++i) {
      solve(i + 1);
    }
    return 0;
  }
  