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;
  set<long long> nums;
  pair<long long, long long> cnt(vector<long long> p) {
    long long a = 0, b = 0;
    for (long long i = 2; i < p.size(); i++) {
      if (p[i - 2] < p[i - 1] && p[i - 1] > p[i]) {
        a++;
      }
      if (p[i - 2] > p[i - 1] and p[i - 1] < p[i]) {
        b++;
      }
    }
    return {a, b};
  }
  void rev(vector<long long>& ans) {
    long long n = ans.size();
    for (long long i = 0; i < n; i++) {
      ans[i] = n + 1 - ans[i];
    }
  }
  void solve() {
    long long n, a, b;
    cin >> n >> a >> b;
    long long a0 = a, b0 = b;
    vector<long long> ans;
    for (long long i = 1; i < n + 1; i++) {
      nums.insert(i);
    }
    bool can = 1;
    long long swapd = (a > b);
    if (abs(a - b) > 1 || a + b + 2 > n) {
      can = 0;
      cout << -1 << "\n";
      return;
    }
    long long nxt = 3;
    ans.push_back(1);
    for (long long i = 1; i < n - 1; i += 2) {
      if (a > swapd) {
        ans.push_back(nxt);
        ans.push_back(nxt - 1);
        a--;
      } else {
        ans.push_back(nxt - 1);
        ans.push_back(nxt);
      }
      nxt += 2;
    }
    if (ans.size() < n) {
      ans.push_back(nxt - 1);
    }
    long long ax = cnt(ans).first, bx = cnt(ans).second;
    if (bx < b0) {
      swap(ans[n - 1], ans[n - 2]);
      rev(ans);
    }
    if (ax < a0) {
      swap(ans[n - 1], ans[n - 2]);
    }
    ax = cnt(ans).first, bx = cnt(ans).second;
    for (long long i = 0; i < n; i++) {
      nums.erase(ans[i]);
      cout << ans[i] << " ";
    }
    cout << "\n";
  }
  signed main() {
    long long t = 1000;
    cin >> t;
    srand(time(NULL));
    while (t--) {
      solve();
    }
  }
  