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;
  template <class A, class B>
  ostream& operator<<(ostream& out, const pair<A, B>& a) {
    return out << "(" << a.first << ", " << a.second << ")";
  }
  template <class A>
  ostream& operator<<(ostream& out, const vector<A>& v) {
    out << "[";
    for (int i = 0; i < int(int(v.size())); i++) {
      if (i) out << ", ";
      out << v[i];
    }
    return out << "]";
  }
  mt19937 rnd(time(NULL));
  const int INF = int(1e9);
  const long long INF64 = long long(1e18);
  const int MOD = int(1e9) + 7;
  const long double EPS = 1e-9;
  const long double PI = acos(-1.0);
  int n, a, b;
  bool read() {
    if (scanf("%d%d%d", &n, &a, &b) != 3) return false;
    return true;
  }
  void solve() {
    if (abs(a - b) > 1 || a + b > n - 2) {
      puts("-1");
      return;
    }
    if (a == 0 && b == 0) {
      for (int i = 0; i < int(n); i++) printf("%d ", i + 1);
      puts("");
      return;
    }
    int l = 1, r = n;
    if (a > b) {
      for (int i = 0; i < int(a + b + 1); i++) {
        if (i & 1)
          printf("%d ", r--);
        else
          printf("%d ", l++);
      }
      for (int _ = 0; _ < int(n - (a + b + 1)); _++) printf("%d ", r--);
    } else if (b > a) {
      for (int i = 0; i < int(a + b + 1); i++) {
        if (i & 1)
          printf("%d ", l++);
        else
          printf("%d ", r--);
      }
      for (int _ = 0; _ < int(n - (a + b + 1)); _++) printf("%d ", l++);
    } else {
      printf("%d ", 2);
      for (int i = 0; i < int(a + b); i++) {
        if (i & 1) {
          printf("%d ", l++);
          if (l == 2) l = 3;
        } else
          printf("%d ", r--);
      }
      for (int _ = 0; _ < int(n - (a + b + 1)); _++) {
        printf("%d ", l++);
        if (l == 2) l = 3;
      }
    }
    puts("");
  }
  int main() {
    cerr.precision(15);
    cout.precision(15);
    cerr << fixed;
    cout << fixed;
    int tc;
    scanf("%d", &tc);
    while (tc--) {
      read();
      solve();
    }
  }
  