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;
  struct tri {
    long long t1, t2, t3;
  };
  bool comp(tri a, tri b) { return a.t1 < b.t1; }
  void solvee() {
    int n;
    cin >> n;
    int a, b;
    cin >> a >> b;
    if (a == 0 && b == 0) {
      for (int i = 1; i <= n; i++) cout << i << " ";
      cout << "\n";
      return;
    }
    if (abs(a - b) > 1 || a + b + 2 > n) {
      cout << "-1\n";
    } else {
      int t = 0;
      int p[n + 1];
      if (a > b) {
        for (int i = 1, k = 1; k <= a; i += 2, k++) p[i] = k;
        for (int i = 2, k = n; k > n - a; i += 2, k--) p[i] = k;
        for (int i = a * 2 + 1, k = n - a; i <= n; i++, k--) p[i] = k;
      }
      if (a <= b) {
        for (int i = 2, k = 1; k <= b; i += 2, k++) p[i] = k;
        for (int i = 1, k = n; k > n - b; i += 2, k--) p[i] = k;
        if (a == b)
          for (int i = b * 2 + 1, k = n - b; i <= n; k--, i++) p[i] = k;
        else
          for (int i = b * 2 + 1, k = b + 1; i <= n; k++, i++) p[i] = k;
      }
      for (int i = 1; i <= n; i++) cout << p[i] << " ";
      cout << "\n";
      return;
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie();
    cout.tie();
    int tt = 1;
    cin >> tt;
    while (tt--) {
      solvee();
    }
    return 0;
  }
  