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;
  bool isPowerOfTwo(long long n) {
    if (n == 0) return false;
    return (ceil(log2(n)) == floor(log2(n)));
  }
  long long powermod(long long x, long long y, long long p) {
    long long res = 1;
    x = x % p;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  bool isPerfectSquare(long long x) {
    if (x >= 0) {
      long long sr = sqrt(x);
      return (sr * sr == x);
    }
    return false;
  }
  bool isPrime(long long n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (long long i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  void gourav_sv() {
    long long n, a, b;
    cin >> n >> a >> b;
    if (a + b + 2 > n)
      cout << -1 << "\n";
    else {
      if (a == b or abs(a - b) == 1) {
        if (a == b) {
          cout << 1 << " ";
          for (long long i = 1; i <= a; i++)
            cout << 2 * i + 1 << " " << 2 * i << " ";
          for (long long i = 2 * (a + 1); i <= n; i++) cout << i << " ";
          cout << "\n";
        }
        if (a == b - 1) {
          for (long long i = 1; i <= b; i++)
            cout << 2 * i << " " << 2 * i - 1 << " ";
          for (long long i = 2 * b + 1; i <= n; i++) cout << i << " ";
          cout << "\n";
        }
        if (b == a - 1) {
          for (long long i = 1; i <= (n - 2 * a); i++) cout << i << " ";
          for (long long i = (n - 2 * a + 1); i <= n; i += 2) {
            cout << i + 1 << " " << i << " ";
          }
          cout << "\n";
        }
      } else
        cout << -1 << "\n";
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t = 1;
    cin >> t;
    while (t--) {
      gourav_sv();
    }
  }
  