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>
  namespace infinities {
  const long long maxn = 1e6 + 7;
  const long long INF = 1e9 + 7;
  using namespace std;
  long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 1) + (x << 3) + (ch ^ 48);
      ch = getchar();
    }
    return x * f;
  }
  long long t, n, a, b, p[maxn];
  void print() {
    for (register long long i = 1; i <= n; i++) cout << p[i] << " ";
    cout << "\n";
  }
  signed main() {
    t = read();
    while (t--) {
      n = read(), a = read(), b = read();
      if (abs(a - b) >= 2) {
        cout << "-1\n";
        continue;
      }
      register long long cnt = 0;
      if (a > b) {
        for (register long long i = 1; i <= b + 1; i++) {
          p[++cnt] = n - a * 2 + i, p[++cnt] = n - a * 2 + i + b + 1;
        }
        if (n - cnt < 1) {
          puts("-1");
          continue;
        }
        for (register long long i = cnt + 1; i <= n; i++) p[i] = n - i + 1;
        print();
        continue;
      }
      if (a < b) {
        for (register long long i = 1; i <= a + 1; i++) {
          p[++cnt] = i + b, p[++cnt] = i;
        }
        if (n - cnt < 1) {
          puts("-1");
          continue;
        }
        for (register long long i = cnt + 1; i <= n; i++) p[i] = i;
        print();
        continue;
      }
      if (a == b) {
        for (register long long i = 1; i <= a + 1; i++)
          p[++cnt] = i, p[++cnt] = i + b + 1;
        if (n - cnt < 0) {
          puts("-1");
          continue;
        }
        for (register long long i = cnt + 1; i <= n; i++) p[++cnt] = i;
        print();
        continue;
      }
    }
    return 0;
  }
  }  // namespace infinities
  signed main() { return infinities::main(); }
  