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;
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
      int n = 10, a, b;
      cin >> n >> a >> b;
      if (abs(a - b) > 1 || (a + b + 2) > n)
        cout << -1 << '\n';
      else {
        vector<int> ans(n);
        iota(ans.begin(), ans.end(), 1);
        int i = a >= b ? n - 1 : 0;
        int j = a >= b ? -2 : 2;
        int k = a >= b ? -1 : 1;
        int x = max(a, b);
        if (a == b) i--;
        while (x--) {
          swap(ans[i], ans[i + k]);
          i += j;
        }
        int g = 0, d = 0;
        for (int i = 1; i < n - 1; ++i) {
          if (ans[i - 1] > ans[i] && ans[i] < ans[i + 1]) g++;
          if (ans[i - 1] < ans[i] && ans[i + 1] < ans[i]) d++;
        }
        for (auto &i : ans) {
          cout << i << " ";
        }
        cout << '\n';
      }
    }
    return 0;
  }
  