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 long long M = 1000000007;
  const long long MM = 998244353;
  const long long INF = 9e18;
  long long expm(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = (res % M * a % M) % M;
      a = (a % M * a % M) % M;
      b >>= 1;
    }
    return res;
  }
  signed main() {
    ios::sync_with_stdio(NULL), cin.tie(NULL), cout.tie(NULL);
    auto start = chrono::high_resolution_clock::now();
    long long tt = 1;
    cin >> tt;
    for (long long test = 1; test <= tt; test++) {
      long long n, a, b;
      cin >> n >> a >> b;
      long long x = abs(a - b);
      if (x > 1 || max(a, b) > n / 2 || (a + b) >= n - 1) {
        cout << -1 << "\n";
        continue;
      }
      if (a == b) {
        long long p = 1, q = n;
        for (long long i = 1; i <= a; i++) {
          cout << p << " " << q << " ";
          p++;
          q--;
        }
        while (p <= q) {
          cout << p << " ";
          p++;
        }
        cout << "\n";
      } else if (a > b) {
        long long p = 1, q = n;
        for (long long i = 1; i <= a; i++) {
          cout << p << " " << q << " ";
          p++;
          q--;
        }
        while (q >= p) {
          cout << q << " ";
          q--;
        }
        cout << "\n";
      } else {
        long long p = 1, q = n;
        for (long long i = 1; i <= b; i++) {
          cout << q << " " << p << " ";
          p++;
          q--;
        }
        while (p <= q) {
          cout << p << " ";
          p++;
        }
        cout << "\n";
      }
    }
    cout << "\n";
    auto stop = chrono::high_resolution_clock::now();
    auto duration = chrono::duration_cast<chrono::microseconds>(stop - start);
    cerr << duration.count() / 1000.0 << " ms\n";
    return 0;
  }
  