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 = 998244353;
  const long long inf = 1e18;
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  template <class T>
  void ckadd(T& a, const T& b) {
    a = (a + b) % mod;
  }
  template <class T>
  void cksub(T& a, const T& b) {
    a = (a - b + mod) % mod;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      long long n, a, b;
      cin >> n >> a >> b;
      if (abs(a - b) > 1) {
        cout << -1 << "\n";
        continue;
      }
      vector<long long> ans;
      if (b > a) ans.push_back(n);
      for (long long i = 0; i < int(n - 1); ++i) {
        ans.push_back(i + 1);
      }
      if (a >= b) ans.push_back(n);
      int i = 1;
      if (b > a) i++;
      if (b > a && i < n) b--;
      while (i < n - 1 && a) {
        swap(ans[i], ans[i + 1]);
        a--;
        i += 2;
        if (i < n) b--;
        if (b == -1) {
          i -= 2;
          swap(ans[i], ans[i + 1]);
          for (int k = 0; i + k < n - 1 - k; k++) {
            swap(ans[i + k], ans[n - 1 - k]);
          }
          b++;
          break;
        }
      }
      if (a == 0 && b == 0) {
        for (auto e : ans) {
          cout << e << " ";
        }
        cout << "\n";
      } else
        cout << -1 << "\n";
    }
  }
  