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;
  bool isrange(int second, int first, int n, int m) {
    if (0 <= second && second < n && 0 <= first && first < m) return true;
    return false;
  }
  int dy[4] = {1, 0, -1, 0}, dx[4] = {0, 1, 0, -1},
      ddy[8] = {1, 0, -1, 0, 1, 1, -1, -1}, ddx[8] = {0, 1, 0, -1, 1, -1, 1, -1};
  void solve(int tc) {
    int n, a, b;
    cin >> n >> a >> b;
    if (abs(a - b) > 1 || a + b > n - 2)
      cout << "-1\n";
    else {
      if (a >= b) {
        int res = n - (a + b + 2);
        for (int e = 1; e <= res; e++) cout << e << " ";
        int lx = res + 1, rx = n;
        for (int e = 0; e < n - res; e++) {
          if (e % 2 == 0) {
            cout << (lx++) << " ";
          } else {
            cout << (rx--) << " ";
          }
        }
      } else {
        int res = n - (a + b + 2);
        for (int e = n; e > n - res; e--) cout << e << " ";
        int rx = n - res, lx = 1;
        for (int e = 0; e < n - res; e++) {
          if (e % 2 == 0) {
            cout << (rx--) << " ";
          } else {
            cout << (lx++) << " ";
          }
        }
      }
      cout << "\n";
    }
  }
  int main(void) {
    ios_base ::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int tc = 1;
    cin >> tc;
    for (int test_number = 1; test_number <= tc; test_number++) {
      solve(test_number);
    }
    return 0;
  }
  