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;
  void solve();
  int main() {
    int t = 1;
    while (t--) {
      solve();
      cout << endl;
    }
    return 0;
  }
  void solve() {
    int n, a, b;
    cin >> n >> a >> b;
    if (abs(a - b) <= 1 && a + b + 2 <= n) {
      if (a <= b) {
        int i, j, k = min(a, b);
        if (a < b) {
          cout << 2 << " " << 1 << " ";
          i = 4, j = 3;
        } else {
          cout << 1 << " ";
          i = 3, j = 2;
        }
        for (; max(i, j) < n && k > 0; i += 2, j += 2, k--)
          cout << i << " " << j << " ";
        for (int t = min(i, j); t <= n; t++) cout << t << " ";
      } else {
        int i = n - 1, j = n, k = max(a, b);
        for (; min(i, j) > 0 && k > 0; i -= 2, j -= 2, k--)
          cout << i << " " << j << " ";
        for (int t = max(i, j); t > 0; t--) cout << t << " ";
      }
    } else {
      cout << -1;
    }
  }
  