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>
  const int N = 1e5 + 10;
  using namespace std;
  int n, m, a, b, r[N];
  void solve() {
    cin >> n >> a >> b;
    if (abs(a - b) > 1 || a + b > n - 2) {
      cout << -1 << endl;
      return;
    }
    if (!(a + b)) {
      for (int i = 1; i <= n; ++i) cout << i << " ";
      cout << endl;
      return;
    }
    if (a > b) {
      r[0] = 1, r[n - 1] = 2;
      for (int i = 0; i < a - 1; ++i) {
        r[2 * i + 1] = n - i, r[2 * (i + 1)] = i + 3;
      }
      r[2 * a - 1] = n - a + 1;
      for (int i = a * 2; i < n - 1; ++i) {
        r[i] = n - a - (i - a * 2);
      }
    }
    if (a == b) {
      r[0] = 1, r[n - 1] = n;
      for (int i = 0; i < a - 1; ++i) {
        r[2 * i + 1] = n - i - 1, r[2 * (i + 1)] = i + 2;
      }
      r[2 * a - 1] = n - a;
      for (int i = a * 2; i < n - 1; ++i) {
        r[i] = n - a - (i - a * 2) - 1;
      }
    }
    if (a < b) {
      r[0] = n - 1, r[n - 1] = n;
      for (int i = 0; i < b - 1; ++i) {
        r[2 * i + 1] = i + 1, r[2 * (i + 1)] = n - i - 2;
      }
      r[2 * b - 1] = b;
      for (int i = b * 2; i < n - 1; ++i) {
        r[i] = b + i - (b * 2) + 1;
      }
    }
    for (int i = 0; i < n; ++i) cout << r[i] << " ";
    cout << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(2);
    int t = 1;
    cin >> t;
    while (t--) solve();
  }
  