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 mod = 998244353;
  const int mx = 1e6;
  int posx[] = {1, -1, 0, 0};
  int posy[] = {0, 0, 1, -1};
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1, n, k, r1, r2, c1, c2, d1, d2, a, b, c, d;
    cin >> t;
    while (t--) {
      cin >> n >> a >> b;
      vector<int> ans(n + 1);
      set<int> s;
      for (int i = 1; i <= n; i++) s.insert(i);
      int pos = 1, f = 0, aa = a, bb = b;
      while (pos < n && (a > 0 || b > 0)) {
        int it = *s.begin();
        if (aa > bb) {
          if (pos & 1)
            it = *s.begin(), b--;
          else
            it = *s.rbegin(), a--;
          f = 0;
          b += (pos == 1);
        } else {
          if (pos & 1)
            it = *s.rbegin(), a--;
          else
            it = *s.begin(), b--;
          f = 1;
          a += (pos == 1);
        }
        ans[pos++] = it;
        s.erase(it);
      }
      if (aa == bb) f = 0;
      if (f == 0) {
        while (!s.empty()) {
          ans[pos++] = *s.rbegin();
          s.erase(*s.rbegin());
        }
      } else {
        while (!s.empty()) {
          ans[pos++] = *s.begin();
          s.erase(*s.begin());
        }
      }
      bool bad = (a > 0 || b > 0);
      a = aa, b = bb;
      for (int i = 2; i < n; i++) {
        a -= (ans[i] > ans[i - 1] && ans[i] > ans[i + 1]);
        b -= (ans[i] < ans[i - 1] && ans[i] < ans[i + 1]);
      }
      bad |= (a != 0 | b != 0);
      if (bad) {
        cout << "-1";
        cout << endl;
        continue;
      }
      for (int i = 1; i <= n; i++) cout << ans[i] << " ";
      cout << endl;
    }
    return 0;
  }
  