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 INF = 1e18;
  const long long MOD = 1e9 + 7;
  const int inf = 1e9;
  const int N = 3e5 + 5;
  bool flag;
  template <typename T>
  istream& operator>>(istream& is, vector<T>& v) {
    for (auto& i : v) is >> i;
    return is;
  }
  template <typename T>
  void mul(T& a, T b) {
    a += b;
    if (b < (T)0) a += MOD;
    if (a >= (T)MOD) a -= MOD;
  }
  void solve() {
    int n, a, b;
    cin >> n >> a >> b;
    if (abs(a - b) > 1 or a + b + 2 > n) {
      cout << -1;
      return;
    }
    if (a == 0 and b == 0) {
      for (int i = 1; i <= n; i++) cout << i << ' ';
      return;
    }
    vector<int> ans;
    int l = 1, r = n;
    if (a > b) {
      ans.push_back(l++);
      for (int i = 1; i <= a + b; i++) {
        if (i % 2)
          ans.push_back(r--);
        else
          ans.push_back(l++);
      }
      for (int i = a + b + 1; i < n; i++) ans.push_back(r--);
    } else {
      ans.push_back(r--);
      for (int i = 1; i <= a + b; i++) {
        if (i % 2 == 0)
          ans.push_back(r--);
        else
          ans.push_back(l++);
      }
      for (int i = a + b + 1; i < n; i++) ans.push_back(a == b ? r-- : l++);
    }
    for (auto now : ans) cout << now << ' ';
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(12);
    int tt = 1;
    cin >> tt;
    while (tt--) {
      flag = 0;
      solve();
      cout << '\n';
    }
    return 0;
  }
  