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 + 100;
  int z[N], ans[N];
  int n, m;
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int a, b, f = 0;
      cin >> n >> a >> b;
      if (a == b) f = 1;
      if (abs(a - b) > 1 || a + b > n - 2) {
        cout << -1 << endl;
        continue;
      } else if (a >= b)
        for (int i = 1; i <= n; i++) z[i] = i;
      else
        for (int i = 1; i <= n; i++) z[i] = n - i + 1;
      ans[1] = z[1];
      int i = 2, j = n, k = 2, sum = a + b;
      for (int p = 1; p <= sum; k++, p++) {
        if (p & 1)
          ans[k] = z[j--], b--;
        else
          ans[k] = z[i++], a--;
      }
      if (f)
        while (k <= n) ans[k++] = z[i++];
      else
        while (k <= n) ans[k++] = z[j--];
      for (int p = 1; p <= n; p++) cout << ans[p] << ' ';
      cout << endl;
    }
  }
  