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;
  const int N = 1e5 + 10;
  int arr[N];
  int main(void) {
    int T;
    cin >> T;
    while (T--) {
      int n, a, b;
      cin >> n >> a >> b;
      bool flag = true;
      for (int i = 1; i <= n; i++) arr[i] = i;
      int maxv = (n - 2) / 2 + (n & 1);
      if (a > maxv || b > maxv)
        flag = false;
      else if (abs(a - b) > 1)
        flag = false;
      else if ((n & 1) && a == maxv && b == maxv)
        flag = false;
      else if (a == 0 && b == 0) {
        for (int i = 1; i <= n; i++) arr[i] = i;
      } else {
        if (a == b) {
          int l = 1, r = n;
          for (int i = 1; i <= b * 2 + 1; i += 2) arr[i] = r--;
          for (int i = 2; i <= a * 2; i += 2) arr[i] = l++;
          for (int i = 2 * b + 2; i <= n; i++) arr[i] = r--;
        } else if (a == b + 1) {
          int l = 1, r = n;
          for (int i = 1; i <= a * 2; i += 2) arr[i] = l++;
          for (int i = 2; i <= a * 2; i += 2) arr[i] = r--;
          for (int i = 2 * a + 1; i <= n; i++) arr[i] = r--;
        } else {
          int l = 1, r = n;
          for (int i = 1; i <= b * 2; i += 2) arr[i] = r--;
          for (int i = 2; i <= b * 2; i += 2) arr[i] = l++;
          for (int i = 2 * b + 1; i <= n; i++) arr[i] = l++;
        }
      }
      if (flag) {
        for (int i = 1; i <= n; i++) cout << arr[i] << " ";
        cout << endl;
      } else
        cout << "-1" << endl;
    }
    return 0;
  }
  