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;
  const long long mod = 1000000007;
  const long long inf = 1000000000000000000;
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n, a, b;
      cin >> n >> a >> b;
      vector<long long> ans;
      vector<long long> p;
      for (long long i = 1; i < n + 1; i += 1) p.push_back(i);
      if (abs(a - b) > 1 || (a + b) > (n - 2))
        cout << "-1"
             << "\n";
      else {
        long long x = 0, y = 0;
        if (a > b) {
          while (a != 0) {
            ans.push_back(p[0 + y]);
            ans.push_back(p[n - 1 - x]);
            y++;
            x++;
            a--;
            b--;
          }
          for (long long i = n - 1 - x; i >= y; i--) ans.push_back(p[i]);
        } else if (a < b) {
          while (b != 0) {
            ans.push_back(p[n - 1 - x]);
            ans.push_back(p[0 + y]);
            y++;
            x++;
            a--;
            b--;
          }
          for (long long i = y; i <= n - 1 - x; i++) ans.push_back(p[i]);
        } else {
          ans.push_back(1);
          y++;
          while (a != 0) {
            ans.push_back(p[n - 1 - x]);
            ans.push_back(p[0 + y]);
            y++;
            x++;
            a--;
            b--;
          }
          for (long long i = y; i <= n - 1 - x; i++) ans.push_back(p[i]);
        }
        for (long long i = 0; i < n; i += 1) cout << ans[i] << " ";
        cout << "\n";
      }
    }
  }
  