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>
  const int N = 2e5 + 5;
  int T, n, a, b, ans[N];
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d %d %d", &n, &a, &b);
      if (abs(b - a) > 1 || a + b + 2 > n)
        puts("-1");
      else {
        if (a < b) {
          int i = 1, j = n;
          for (int k = 1; k <= a + 1; k++, j--) ans[k * 2 - 1] = j;
          for (int k = 1; k <= b; k++, i++) ans[k * 2] = i;
          for (int k = a + b + 2; k <= n; k++, i++) ans[k] = i;
        } else if (a > b) {
          int i = 1, j = n;
          for (int k = 1; k <= a; k++, j--) ans[k * 2] = j;
          for (int k = 1; k <= b + 1; k++, i++) ans[k * 2 - 1] = i;
          for (int k = a + b + 2; k <= n; k++, j--) ans[k] = j;
        } else if (a == b && (a == 0 || b == 0)) {
          for (int i = 1; i <= n; i++) ans[i] = i;
        } else {
          int j = n;
          for (int i = 1; i <= b + 1; i++) ans[i * 2 - 1] = i;
          for (int i = 1; i <= b; i++, j--) ans[i * 2] = j;
          j--;
          for (int i = 2 * b + 2; i <= n; i++, j--) ans[i] = n - j + 1;
        }
        for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
        puts("");
      }
    }
    return 0;
  }
  