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>
  char ibuf[1 << 15], *p1, *p2;
  struct FastIO {
    inline int rint() {
      register char c, f = 0;
      while ((c = (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, 1 << 15, stdin),
                                p1 == p2)
                       ? EOF
                       : *p1++)) < 48 ||
             c > 57)
        f |= c == '-';
      register int unsigned a = c & 15;
      while ((c = (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, 1 << 15, stdin),
                                p1 == p2)
                       ? EOF
                       : *p1++)) >= 48 &&
             c <= 57)
        a = a * 10 + (c & 15);
      return f ? ~a + 1 : a;
    }
    inline long long rll() {
      register char c, f = 0;
      while ((c = (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, 1 << 15, stdin),
                                p1 == p2)
                       ? EOF
                       : *p1++)) < 48 ||
             c > 57)
        f |= c == '-';
      register unsigned long long a = c & 15;
      while ((c = (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, 1 << 15, stdin),
                                p1 == p2)
                       ? EOF
                       : *p1++)) >= 48 &&
             c <= 57)
        a = a * 10 + (c & 15);
      return f ? ~a + 1 : a;
    }
    inline operator int() { return rint(); }
  } g90;
  const int N = 100100;
  signed main() {
    int T = 1;
    T = g90;
    for (; T--;) {
      static int p[N];
      int n = g90, a = g90, b = g90;
      if (a < b - 1 || a > b + 1 || a + b > n - 2) {
        puts("-1");
        continue;
      } else if (a == b + 1) {
        for (int i = 1; i <= n - 1 - a - b; ++i) {
          p[i] = i;
        }
        for (int i = n - a - b, j = n; i <= n; i += 2, j--) {
          p[i] = j;
        }
        for (int i = n - a - b + 1, j = n - a - b; i <= n; i += 2, j++) {
          p[i] = j;
        }
      } else {
        for (int i = 1; i <= n - 1 - a - b; ++i) {
          p[i] = n - i + 1;
        }
        for (int i = n - a - b, j = 1; i <= n; i += 2, j++) {
          p[i] = j;
        }
        for (int i = n - a - b + 1, j = n - (n - a - b) + 1; i <= n;
             i += 2, j--) {
          p[i] = j;
        }
      }
      for (int i = 1; i <= n; ++i) {
        printf("%d ", p[i]);
      }
      puts("");
    }
  }
  