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 int mod = 100000007;
  const int mini = -2147483648;
  const int maxi = 2147483647;
  void _print(long long int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  void solve() {
    long long int n, a, b;
    cin >> n >> a >> b;
    vector<long long int> v(n);
    if ((a + b) > (n - 2) || abs(a - b) > 1) {
      cout << -1;
      return;
    }
    for (long long int i = 0; i < n; i++) {
      v[i] = i + 1;
    }
    long long int maxx = max(a, b);
    if (a == b) {
      long long int cnt = 0;
      for (long long int i = 1; i < n - 1; i++) {
        if (cnt == a) break;
        swap(v[i], v[i + 1]);
        cnt++;
        i++;
      }
    } else {
      long long int cnt = 0;
      for (long long int i = 0; i < n; i++) {
        if (maxx > cnt) {
          swap(v[i], v[i + 1]);
          cnt++;
          i++;
        } else
          break;
      }
      if (maxx == a) {
        for (long long int i = 0; i < n; i++) {
          v[i] = n - v[i] + 1;
        }
      }
    }
    for (auto &i : v) {
      cout << i << " ";
    }
  }
  signed main() {
    freopen("Error.txt", "w", stderr);
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    auto start = chrono::high_resolution_clock::now();
    long long int t = 1;
    cin >> t;
    while (t--) {
      solve();
      cout << "\n";
    }
    auto stop = chrono::high_resolution_clock::now();
    auto duration = chrono::duration_cast<chrono::microseconds>(stop - start);
    cerr << "\n"
         << "Time : " << duration.count() / 1e3 << " ms\n";
  }
  