#include <bits/stdc++.h>
using namespace std;
int main() {
  int t, a, b, n, i;
  cin >> t;
  while (t--) {
    cin >> n >> a >> b;
    if (a - b > 1 || a - b < -1 || n - 2 < a + b) {
      cout << "-1" << endl;
      continue;
    }
    if (a == 0 && b == 0) {
      cout << "1";
      for (int j = 2; j <= n; j++) cout << " " << j;
      cout << endl;
      continue;
    }
    if (a == b) {
      i = 0;
      cout << a + 1;
      while (i < a) {
        i++;
        cout << " " << i;
        if (i == 1)
          for (int j = a + b + 2; j <= n; j++) cout << " " << j;
        else
          cout << " " << a + b - i + 3;
      }
      cout << " " << a + 2 << endl;
    } else if (a > b) {
      i = 0;
      cout << a + 1;
      while (i < a) {
        i++;
        if (i == 1)
          for (int j = a + b + 2; j <= n; j++) cout << " " << j;
        else
          cout << " " << a + b - i + 3;
        cout << " " << i;
      }
      cout << endl;
    } else {
      i = 0;
      cout << b + 1;
      while (i < b) {
        i++;
        cout << " " << i;
        if (i == 1)
          for (int j = a + b + 2; j <= n; j++) cout << " " << j;
        else
          cout << " " << a + b - i + 3;
      }
      cout << endl;
    }
  }
}

### Prompt

Develop a solution in CPP 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;
int main() {
  int t, a, b, n, i;
  cin >> t;
  while (t--) {
    cin >> n >> a >> b;
    if (a - b > 1 || a - b < -1 || n - 2 < a + b) {
      cout << "-1" << endl;
      continue;
    }
    if (a == 0 && b == 0) {
      cout << "1";
      for (int j = 2; j <= n; j++) cout << " " << j;
      cout << endl;
      continue;
    }
    if (a == b) {
      i = 0;
      cout << a + 1;
      while (i < a) {
        i++;
        cout << " " << i;
        if (i == 1)
          for (int j = a + b + 2; j <= n; j++) cout << " " << j;
        else
          cout << " " << a + b - i + 3;
      }
      cout << " " << a + 2 << endl;
    } else if (a > b) {
      i = 0;
      cout << a + 1;
      while (i < a) {
        i++;
        if (i == 1)
          for (int j = a + b + 2; j <= n; j++) cout << " " << j;
        else
          cout << " " << a + b - i + 3;
        cout << " " << i;
      }
      cout << endl;
    } else {
      i = 0;
      cout << b + 1;
      while (i < b) {
        i++;
        cout << " " << i;
        if (i == 1)
          for (int j = a + b + 2; j <= n; j++) cout << " " << j;
        else
          cout << " " << a + b - i + 3;
      }
      cout << endl;
    }
  }
}

```