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 read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - 48;
      ch = getchar();
    }
    return x * f;
  }
  const long long maxn = 2e5 + 1000;
  long long t, n, m, a, b, si[maxn];
  signed main() {
    t = read();
    while (t--) {
      n = read();
      a = read();
      b = read();
      long long dis = abs(a - b);
      if (a + b + 2 > n || dis > 1)
        cout << -1 << "\n";
      else {
        if (a > b) {
          for (long long i = 1; i <= a; ++i) cout << i << ' ' << n - i + 1 << ' ';
          long long l = a + 1, r = n + 1 - l;
          for (long long i = r; i >= l; --i) cout << i << ' ';
          cout << "\n";
        } else if (a == b) {
          for (long long i = 1; i <= a; ++i) cout << i << ' ' << n - i + 1 << ' ';
          long long l = a + 1, r = n + 1 - l;
          for (long long i = l; i <= r; ++i) cout << i << ' ';
          cout << "\n";
        } else {
          for (long long i = 1; i <= b; ++i) cout << n - i + 1 << ' ' << i << ' ';
          long long l = b + 1, r = n + 1 - l;
          for (long long i = l; i <= r; ++i) cout << i << ' ';
          cout << "\n";
        }
      }
    }
  }
  