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;
  const int N = 4e5 + 50;
  const int M = 5e3 + 50;
  int n, m, u, k, l, r, t, x, y;
  int cnt, cnt1, cnt2, ans;
  int mn = INT_MAX;
  int mx = INT_MIN;
  const int fx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
  const int fy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
  const int dx[] = {0, 1, 0, -1};
  const int dy[] = {1, 0, -1, 0};
  const int MOD = (1ll << 30);
  const int mod = 1e9 + 7;
  bool cmp(pair<int, int> a, pair<int, int> b) {
    if (abs(a.second) == abs(b.second)) {
      return abs(a.first) < abs(b.first);
    }
    return abs(a.second) < abs(b.second);
  }
  bool cmp1(const pair<string, pair<int, int>> &a,
            const pair<string, pair<int, int>> &b) {
    return abs(a.second.first) < abs(b.second.first);
  }
  bool isPowerOfTwo(long long n) { return (ceil(log2(n)) == floor(log2(n))); }
  int makedigit(string n, int x) {
    int res = 0;
    for (int i = 0; i < x; i++) {
      res = res * 10 + (n[i] - '0');
    }
    return res;
  }
  string inttostrng(int a) {
    ostringstream temp;
    string str;
    temp << a;
    return temp.str();
  }
  void clearr(int vis[N], int n) {
    for (int i = 0; i < n; i++) {
      vis[i] = 0;
    }
  }
  long long pow(long long a, long long b, long long m) {
    long long ans = 1;
    while (b) {
      if (b & 1) {
        ans = (ans * a) % m;
      }
      b /= 2;
      a = (a * a) % m;
    }
    return ans;
  }
  bool is_prime[N];
  void prime() {
    is_prime[0] = is_prime[1] = true;
    for (int i = 2; i * i <= N; i++) {
      if (is_prime[i] == false) {
        for (int j = i * i; j <= N; j += i) {
          is_prime[j] = true;
        }
      }
    }
  }
  bool is_palidrome(string second) {
    for (int i = 0; i < second.size() / 2; i++) {
      if (second[i] != second[second.size() - i - 1]) {
        return false;
      }
    }
    return true;
  }
  long long sum(long long x) {
    if (x < 0) {
      return -1;
    }
    long long second = 0;
    while (x) {
      int rem = x % 10;
      second += rem;
      x /= 10;
    }
    return second;
  }
  long long poww(long long x, long long y) {
    long long pw = 1;
    for (int i = 1; i <= y; i++) {
      pw *= x;
    }
    return pw;
  }
  long long pp(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) {
        ans = (ans * a);
      }
      b >>= 1;
      a = (a * a);
    }
    return ans;
  }
  void dfs(int x, int y, char c[200][200], int n) {
    if (x <= 0 or x > 2 or y <= 0 or y > n) {
      return;
    }
    if (c[x][y] == '1' or c[x][y] == '.') {
      return;
    }
    if (c[x][y] == '0') {
      c[x][y] = '.';
    }
    dfs(x, y + 1, c, n);
    dfs(x + 1, y, c, n);
    dfs(x + 1, y + 1, c, n);
    dfs(x - 1, y, c, n);
    dfs(x - 1, y + 1, c, n);
    dfs(x, y + 1, c, n);
  }
  int getPairsCount(int arr[], int n, int diff) {
    int ret = 0;
    map<int, int> m;
    for (int i = 1; i <= n; i++) {
      if (m.find(diff - arr[i]) != m.end()) {
        ret += m[diff - arr[i]];
      }
      m[arr[i]]++;
    }
    return ret;
  }
  bool chk(int n) {
    for (int i = 2; i * i <= n; ++i) {
      if (n % i == 0) return false;
    }
    if (n == 1) return false;
    return true;
  }
  void solve() {
    cin >> n >> l >> r;
    vector<int> v(n);
    if (abs(l - r) > 1) {
      cout << -1 << "\n";
      return;
    } else if ((l + r) > n - 2) {
      cout << -1 << "\n";
      return;
    }
    if (l == 0 and r == 0) {
      for (int i = 0; i < n; i++) {
        v[i] = i + 1;
      }
    } else if (l == r) {
      int second = 1, e = n, a = l, b = r;
      for (int i = 0; i < n; i++) {
        if (b == 0) {
          v[i] = e;
          e--;
          continue;
        }
        if (i % 2 == 1) {
          v[i] = second;
          second++;
        } else {
          v[i] = e;
          e--;
        }
        if (i > 1 and i < n - 1 and b != 0) {
          if (v[i - 1] < v[i] and v[i - 1] < v[i - 2]) {
            b--;
          }
        }
      }
    } else if (l > r and r != 0) {
      int second = 1, e = n, a = l, b = r;
      for (int i = 0; i < n; i++) {
        if (b == 0) {
          v[i] = e;
          e--;
          continue;
        }
        if (i % 2 == 0) {
          v[i] = second;
          second++;
        } else {
          v[i] = e;
          e--;
        }
        if (i > 1 and i < n - 1 and b != 0) {
          if (v[i - 1] > v[i] and v[i - 1] > v[i - 2]) {
            a--;
          }
          if (v[i - 1] < v[i] and v[i - 1] < v[i - 2]) {
            b--;
          }
        }
      }
    } else if (l > r and r == 0) {
      for (int i = 0; i < n; i++) {
        v[i] = i + 1;
      }
      swap(v[n - 1], v[n - 2]);
    } else if (l < r and l != 0) {
      int second = 1, e = n, a = l, b = r;
      for (int i = 0; i < n; i++) {
        if (a == 0) {
          v[i] = second;
          second++;
          continue;
        }
        if (i % 2 == 1) {
          v[i] = second;
          second++;
        } else {
          v[i] = e;
          e--;
        }
        if (i > 1 and i < n - 1 and a != 0) {
          if (v[i - 1] > v[i] and v[i - 1] > v[i - 2]) {
            a--;
          }
          if (v[i - 1] < v[i] and v[i - 1] < v[i - 2]) {
            b--;
          }
        }
      }
    } else if (l < r and l == 0) {
      for (int i = 0; i < n; i++) {
        v[i] = i + 1;
      }
      swap(v[0], v[1]);
    }
    int a = 0, b = 0;
    for (int i = 1; i < n - 1; i++) {
      if (v[i - 1] < v[i] and v[i] > v[i + 1]) {
        a++;
      }
      if (v[i - 1] > v[i] and v[i] < v[i + 1]) {
        b++;
      }
    }
    if (a != l or b != r) {
      cout << -1 << "\n";
    } else {
      for (int i = 0; i < n; i++) {
        cout << v[i] << " ";
      }
      cout << "\n";
    }
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    ios_base::sync_with_stdio(0);
    cin >> t;
    while (t--) {
      solve();
    }
  }
  