#include <bits/stdc++.h>
using namespace std;
void code() {
  long long n, a, b;
  cin >> n >> a >> b;
  if (a + b > n - 2 or abs(a - b) > 1)
    cout << -1 << endl;
  else {
    if (a < b) {
      long long tp = a + b + 1;
      long long i = 1;
      long long j = a + b + 1;
      bool flag = 1;
      while (tp--) {
        if (flag) {
          cout << j << " ";
          j--;
          flag = 0;
        } else {
          cout << i << " ";
          i++;
          flag = 1;
        }
      }
      for (long long k = a + b + 2; k <= n; k++) {
        cout << k << " ";
      }
      cout << endl;
    } else if (a >= b) {
      long long tp = a + b + 1;
      long long i = n - tp + 1;
      long long j = n;
      bool flag = 1;
      for (long long k = 1; k < i; k++) {
        cout << k << " ";
      }
      while (tp--) {
        if (flag) {
          cout << j << " ";
          j--;
          flag = 0;
        } else {
          cout << i << " ";
          i++;
          flag = 1;
        }
      }
      cout << endl;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T;
  cin >> T;
  while (T--) code();
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void code() {
  long long n, a, b;
  cin >> n >> a >> b;
  if (a + b > n - 2 or abs(a - b) > 1)
    cout << -1 << endl;
  else {
    if (a < b) {
      long long tp = a + b + 1;
      long long i = 1;
      long long j = a + b + 1;
      bool flag = 1;
      while (tp--) {
        if (flag) {
          cout << j << " ";
          j--;
          flag = 0;
        } else {
          cout << i << " ";
          i++;
          flag = 1;
        }
      }
      for (long long k = a + b + 2; k <= n; k++) {
        cout << k << " ";
      }
      cout << endl;
    } else if (a >= b) {
      long long tp = a + b + 1;
      long long i = n - tp + 1;
      long long j = n;
      bool flag = 1;
      for (long long k = 1; k < i; k++) {
        cout << k << " ";
      }
      while (tp--) {
        if (flag) {
          cout << j << " ";
          j--;
          flag = 0;
        } else {
          cout << i << " ";
          i++;
          flag = 1;
        }
      }
      cout << endl;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T;
  cin >> T;
  while (T--) code();
}

```