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 INF = 9e18;
  const long long N = 2e5 + 5;
  const long long M = 1e9 + 7;
  bool chk(vector<long long> a, long long x, long long y) {
    long long A = 0, B = 0;
    for (long long i = 2; i < a.size(); i++)
      if (a[i] > a[i - 1] && a[i] > a[i + 1])
        A++;
      else if (a[i] < a[i - 1] && a[i] < a[i + 1])
        B++;
    if (A == x && B == y) return 1;
    return 0;
  }
  signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    long long tests;
    cin >> tests;
    while (tests--) {
      vector<long long> a;
      long long n, x, y;
      bool ok = 0;
      cin >> n >> x >> y;
      if (abs(x - y) > 1 || x + y > n - 2) {
        cout << "-1\n";
        continue;
      }
      if (x >= y) {
        long long l = 1;
        long long r = n;
        while (l <= r) {
          a.push_back(l++);
          if (l > r) break;
          a.push_back(r--);
        }
        long long idx = 1 + 2 * y;
        if (x > y) idx++;
        sort(a.begin() + idx, a.begin() + n);
        if (x > y) reverse(a.begin() + idx, a.begin() + n);
      } else {
        long long l = 1;
        long long r = n;
        while (l <= r) {
          a.push_back(r--);
          if (l > r) break;
          a.push_back(l++);
        }
        long long idx = 2 * y;
        sort(a.begin() + idx, a.begin() + n);
      }
      for (auto h : a) cout << h << " ";
      cout << "\n";
    }
  }
  