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;
  vector<long long> prime;
  vector<bool> mark(1e6 + 6, true);
  vector<long long> fact(1e6 + 1, 1);
  vector<long long> inv(1e6 + 1, 1);
  vector<long long> invf(1e6 + 1, 1);
  long long Mod = 1e9 + 7;
  void seive(long long n) {
    long long i, j;
    mark[1] = false;
    prime.push_back(2);
    for (i = 4; i <= n; i += 2) mark[i] = false;
    for (i = 3; i * i <= n; i++) {
      if (mark[i]) {
        for (j = i * i; j <= n; j += 2 * i) {
          mark[j] = false;
        }
      }
    }
    for (i = 3; i <= n; i += 2)
      if (mark[i]) prime.push_back(i);
  }
  long long ck(long long n) {
    long long l = 1, r = n, mid;
    while (l <= r) {
      mid = (l + r) / 2;
      if ((mid * (mid + 1)) / 2 < n)
        l = mid + 1;
      else
        r = mid - 1;
    }
    return l;
  }
  void ft() {
    for (long long i = 2; i <= 1e6; i++) {
      fact[i] = (i * fact[i - 1]) % Mod;
    }
  }
  void modinv() {
    inv[0] = inv[1] = 1;
    for (long long i = 2; i <= 1e6; i++) {
      inv[i] = inv[Mod % i] * (Mod - Mod / i) % Mod;
    }
  }
  void modfactinv() {
    invf[0] = invf[1] = 1;
    for (long long i = 2; i <= 1e6; i++) {
      invf[i] = (invf[i - 1] * inv[i]) % Mod;
    }
  }
  void solve(long long cs) {
    long long n, m, i, z, h, w, e, k = 0, j, a, p, s, b, ans = 2;
    cin >> n >> a >> b;
    if (a == b) {
      k = a * 2 + 2;
    } else if (a < b) {
      k = 2 * (a + 1) + 1;
    } else {
      k = 2 * (b + 1) + 1;
    }
    if (k > n or abs(a - b) > 1)
      cout << "-1";
    else {
      if (a < b) {
        w = (a + 1) * 2 + 1;
        h = 1;
        cout << w-- << " ";
        for (i = 0; i < a + 1; i++) {
          cout << h++ << " " << w-- << " ";
        }
        h = (a + 1) * 2 + 2;
        for (i = 0; i < b - a - 1; i++) {
          cout << h + 1 << " " << h << " " << h + 2 << " ";
          h += 3;
        }
      } else if (b < a) {
        w = n;
        h = n - (b + 1) * 2;
        cout << h++ << " ";
        for (i = 0; i < b + 1; i++) {
          cout << w-- << " " << h++ << " ";
        }
        w = n - (b + 1) * 2 - 1;
        for (i = 0; i < a - b - 1; i++) {
          cout << w - 2 << " " << w << " " << w - 1 << " ";
          w -= 3;
        }
      } else {
        w = k;
        h = 1;
        for (i = 0; i < a + 1; i++) {
          cout << h++ << " " << w-- << " ";
        }
      }
      for (i = k + 1; i <= n; i++) {
        if (a < b)
          cout << i << " ";
        else if (a > b)
          cout << n - i + 1 << " ";
        else
          cout << i << " ";
      }
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t = 1, cs = 1;
    cin >> t;
    while (t--) {
      solve(cs++);
    }
  }
  