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;
  int sgn(int x) {
    if (x < 0) return -1;
    return 1;
  }
  int main() {
    int T;
    cin >> T;
    for (int t = 1; t <= T; t++) {
      int n;
      cin >> n;
      int a, b;
      cin >> a >> b;
      if (abs(a - b) > 1 || a + b > n - 2) {
        cout << -1 << endl;
        continue;
      }
      vector<int> A;
      A.push_back(0);
      if (a >= b) {
        for (int i = 1; i <= a; i++) {
          A.push_back(i);
          A.push_back(-i);
        }
        if (a == b) {
          A.push_back(a + 1);
        }
      } else if (a < b) {
        for (int i = 1; i <= b; i++) {
          A.push_back(-i);
          A.push_back(i);
        }
      }
      int dir = 1;
      if (A.size() >= 2) {
        dir = sgn(A[A.size() - 1] - A[A.size() - 2]);
      }
      while (A.size() < n) {
        A.push_back(A.back() + dir);
      }
      set<int> uniq;
      for (int a : A) {
        uniq.insert(a);
      }
      map<int, int> ids;
      for (int x : uniq) {
        int sz = ids.size();
        ids[x] = sz + 1;
      }
      if (A.size() != n) {
        cout << -1 << endl;
        continue;
      }
      for (int& a : A) {
        a = ids[a];
      }
      for (int i = 1; i <= n - 2; i++) {
        a -= (A[i] > max(A[i - 1], A[i + 1]));
        b -= (A[i] < min(A[i - 1], A[i + 1]));
      }
      if (a != 0 || b != 0) {
        cout << -1 << endl;
        continue;
      }
      for (int& a : A) {
        printf("%d ", a);
      }
      cout << endl;
    }
  }
  