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;
  using ll = long long;
  using ld = long double;
  using vl = vector<ll>;
  using vi = vector<int>;
  using vs = vector<string>;
  using vc = vector<char>;
  using vvl = vector<vl>;
  using P = pair<ll, ll>;
  using vvc = vector<vc>;
  using vd = vector<double>;
  using vp = vector<P>;
  using vb = vector<bool>;
  const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
  const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
  const ll inf = 2e18;
  const ll MOD = 1000000007;
  const ll mod = 998244353;
  const double pi = acos(-1);
  template <typename T1, typename T2>
  ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
    os << p.first << " " << p.second;
    return os;
  }
  template <typename T1, typename T2>
  istream &operator>>(istream &is, pair<T1, T2> &p) {
    is >> p.first >> p.second;
    return is;
  }
  template <typename T>
  ostream &operator<<(ostream &os, const vector<T> &v) {
    for (int i = 0; i < (int)v.size(); i++) {
      os << v[i] << (i + 1 != v.size() ? " " : "");
    }
    return os;
  }
  template <typename T>
  istream &operator>>(istream &is, vector<T> &v) {
    for (T &in : v) {
      is >> in;
    }
    return is;
  }
  void scan() {}
  template <class Head, class... Tail>
  void scan(Head &head, Tail &...tail) {
    cin >> head;
    scan(tail...);
  }
  template <class T>
  void print(const T &t) {
    cout << t << '\n';
  }
  template <class Head, class... Tail>
  void print(const Head &head, const Tail &...tail) {
    cout << head << ' ';
    print(tail...);
  }
  template <class... T>
  void fin(const T &...a) {
    print(a...);
    exit(0);
  }
  template <typename T>
  T sum_(vector<T> a) {
    return accumulate(begin(a), end(a), T(0));
  }
  template <typename T1, typename T2>
  inline bool chmax(T1 &a, T2 b) {
    return a < b && (a = b, true);
  }
  template <typename T1, typename T2>
  inline bool chmin(T1 &a, T2 b) {
    return a > b && (a = b, true);
  }
  void solve();
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(12);
    ll times = 1;
    cin >> times;
    while (times--) {
      solve();
    }
  }
  void solve() {
    ll n, a, b;
    scan(n, a, b);
    ll diff = abs(a - b);
    if (diff > 1) {
      print(-1);
      return;
    }
    if (a + b > (n - 2)) {
      print(-1);
      return;
    }
    set<ll> s;
    for (__typeof(1) i = 1; i < n + 1; i++) s.insert(i);
    ll status = 0;
    if (a < b) status = 1;
    vl ans(n);
    for (__typeof(n - a - b) i = 0; i < n - a - b; i++) {
      if (status)
        ans[i] = n - a - b - i;
      else
        ans[i] = n - (n - a - b - 1) + i;
      s.erase(ans[i]);
    }
    status ^= 1;
    for (__typeof(n - a - b) i = n - a - b; i < n; i++) {
      if (status && a + b != n)
        ans[i] = *(s.begin());
      else if (a + b != n)
        ans[i] = *(--s.end());
      else if (status)
        ans[i] = *(--s.end());
      else
        ans[i] = *(s.begin());
      s.erase(ans[i]);
      status ^= 1;
    }
    print(ans);
  }
  