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>
  #pragma GCC optimize(2, 3, "Ofast")
  using namespace std;
  const long long INF = 1e18 + 100;
  const long long NINF = -INF;
  const int inf = 1e8;
  const int MOD = 998244353;
  const int mod = 1000000007;
  const int ninf = -mod;
  long long mul(long long a, long long b) {
    return ((a % mod) * (b % mod)) % mod;
  }
  long long tri(long long a, long long b, long long c) {
    return mul(mul(a, b), c);
  }
  long long binPow(long long a, long long b, long long mod) {
    if (b == 0) return 1;
    long long ans = binPow(a, b / 2, mod);
    if (b & 1) {
      return tri(ans, ans, a);
    }
    return mul(ans, ans);
  }
  void __print(int x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char *x) { cerr << '\"' << x << '\"'; }
  void __print(const string &x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    int f = 0;
    cerr << '{';
    for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  template <typename T1, typename T2>
  void ckmin(T1 &a, T2 b) {
    if (a > b) a = b;
  }
  template <typename T1, typename T2>
  void ckmax(T1 &a, T2 b) {
    if (a < b) a = b;
  }
  void solve() {
    int n, a, b, i, j, f = 1, t = 1;
    cin >> n >> a >> b;
    if (a + b + 2 > n || abs(a - b) > 1) {
      cout << -1 << "\n";
      return;
    }
    if (a == b) {
      t = 0;
      if (a == 0) {
        for (i = 1; i <= n; i++) cout << i << " ";
        cout << "\n";
        return;
      }
    }
    vector<int> v;
    if (a >= b) {
      int left = 2, right = n - 1;
      cout << 1 << " " << n << " ";
      while (a > 0) {
        if (f) {
          if (a == 1) {
            if (t == 1) {
              cout << right-- << " ";
            } else {
              cout << left++ << " ";
            }
          } else {
            cout << left++ << " ";
          }
          a--;
        } else {
          cout << right-- << " ";
        }
        f = !f;
      }
      if (t == 1) {
        for (i = right; i >= left; i--) cout << i << " ";
        cout << "\n";
        return;
      } else {
        for (i = left; i <= right; i++) cout << i << " ";
        cout << "\n";
        return;
      }
    } else {
      int left = 2, right = n - 1;
      cout << n << " " << 1 << " ";
      while (b > 0) {
        if (f) {
          if (b == 1) {
            cout << left++ << " ";
          } else {
            cout << right-- << " ";
          }
          b--;
        } else {
          cout << left++ << " ";
        }
        f = !f;
      }
      for (i = left; i <= right; i++) cout << i << " ";
      cout << "\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
  }
  