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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const int N = 2e5 + 5;
  const int M = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  const long long LINF = 0x3f3f3f3f3f3f3fll;
  const double pi = acos(-1.0);
  int read() {
    int a = 0;
    int d = 1;
    char c;
    while (c = getchar(), c < '0' || c > '9')
      if (c == '-') d = -1;
    a = a * 10 + c - 48;
    while (c = getchar(), c >= '0' && c <= '9') a = a * 10 + c - 48;
    a *= d;
    return a;
  }
  int aa[N];
  signed main() {
    int t = read();
    while (t--) {
      int n = read();
      int a = read();
      int b = read();
      swap(a, b);
      for (int i = 1; i <= n; i++) aa[i] = i;
      if (abs(a - b) > 1 || a + b > n - 2) {
        puts("-1");
        continue;
      }
      int id;
      if (a < b) {
        swap(aa[n - 1], aa[n]);
        b--;
        id = 2;
      } else if (a > b) {
        swap(aa[1], aa[2]);
        a--;
        id = 3;
      } else
        id = 2;
      for (int i = 1; i <= a; i++) {
        swap(aa[id], aa[id + 1]);
        id += 2;
      }
      for (int i = 1; i <= n; i++) printf("%d ", aa[i]);
      puts("");
    }
    return 0;
  }
  