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;
  long long int M = 1e9 + 7;
  long long int mE(long long int x, long long int n, long long int M) {
    long long int result = 1;
    while (n > 0) {
      if (n % 2 == 1) result = (result * x) % M;
      x = (x * x) % M;
      n = n / 2;
    }
    return result;
  }
  void light_yagami() {
    long long int n, a, b;
    cin >> n >> a >> b;
    long long int aa = a, bb = b;
    vector<long long int> v(n);
    for (int i = 0; i < n; i++) v[i] = i + 1;
    if (a == 0 && b == 0) {
      for (int i = 0; i < n; i++) cout << v[i] << " ";
      cout << endl;
    } else if (n % 2 == 0) {
      if (a >= (n / 2) || b >= (n / 2))
        cout << -1 << endl;
      else if (abs(a - b) > 1)
        cout << -1 << endl;
      else {
        if (a > b) {
          long long int i = n - 1ll;
          while (a--) {
            swap(v[i], v[i - 1]);
            i -= 2ll;
          }
        } else {
          long long int i = 1ll;
          while (b--) {
            swap(v[i], v[i - 1]);
            i += 2ll;
          }
        }
        if (aa == bb) swap(v[n - 1], v[n - 2]);
        for (int i = 0; i < n; i++) cout << v[i] << " ";
        cout << endl;
      }
    } else {
      if (a > ((n / 2)) || b > ((n / 2)))
        cout << -1 << endl;
      else if (a == n / 2 && b == n / 2)
        cout << -1 << endl;
      else if (abs(a - b) > 1)
        cout << -1 << endl;
      else {
        if (a > b) {
          long long int i = n - 1ll;
          while (a--) {
            swap(v[i], v[i - 1]);
            i -= 2ll;
          }
        } else {
          long long int i = 1ll;
          while (b--) {
            swap(v[i], v[i - 1]);
            i += 2ll;
          }
        }
        if (aa == bb) swap(v[n - 1], v[n - 2]);
        for (int i = 0; i < n; i++) cout << v[i] << " ";
        cout << endl;
      }
    }
  }
  int main() {
    int q;
    cin >> q;
    while (q--) {
      light_yagami();
    }
  }
  