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;
  long long int gcd(long long int a, long long int b) {
    return b ? gcd(b, a % b) : a;
  }
  long long int expo(long long int a, long long int b, long long int mod) {
    long long int res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b = b >> 1;
    }
    return res;
  }
  long long int mminvprime(long long int a, long long int b) {
    return expo(a, b - 2, b);
  }
  long long int mod_add(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long int mod_mul(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long int mod_sub(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long int mod_div(long long int a, long long int b, long long int m) {
    a = a % m;
    b = b % m;
    return (mod_mul(a, mminvprime(b, m), m) + m) % m;
  }
  vector<long long int> sieve(int n) {
    int* arr = new int[n + 1]();
    vector<long long int> vect;
    for (int i = 2; i <= n; i++)
      if (arr[i] == 0) {
        vect.push_back(i);
        for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
      }
    return vect;
  }
  void solve() {
    long long int n;
    cin >> n;
    long long int a;
    cin >> a;
    long long int b;
    cin >> b;
    vector<long long int> v;
    if (abs(a - b) > 1 || (n - 2 + 1) / 2 < a || (n - 2 + 1) / 2 < b ||
        (n - 2) < (a + b)) {
      cout << "-1\n";
      return;
    }
    long long int mn = min(a, b);
    long long int points = n;
    if (a > b) {
      long long int i = 1;
      while (mn >= 0 && points > 0) {
        v.push_back(i);
        v.push_back(n - i + 1);
        points -= 2;
        i++;
        mn--;
      }
      long long int start = i - 1;
      while ((n - i + 1) > start && points > 0) {
        v.push_back(n - i + 1);
        points -= 1;
        i++;
      }
    } else if (b > a) {
      long long int i = 1;
      while (mn >= 0 && points > 0) {
        v.push_back(n - i + 1);
        v.push_back(i);
        points -= 2;
        i++;
        mn--;
      }
      long long int end = n - i + 1 + 1;
      while (i < end && points > 0) {
        v.push_back(i);
        points -= 1;
        i++;
      }
    } else {
      long long int i = 1;
      while (mn > 0 && points > 0 && (a + b) > 0) {
        v.push_back(n - i + 1);
        v.push_back(i);
        points -= 2;
        i++;
        mn--;
      }
      long long int start = i - 1;
      while ((n - i + 1) > start && points > 0) {
        v.push_back(n - i + 1);
        points -= 1;
        i++;
      }
    }
    for (auto it : v) {
      cout << it << ' ';
    }
    cout << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    long long int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  