#include <bits/stdc++.h>
using namespace std;
void swap(int *a, int *b) {
  int tmp = *a;
  *a = *b;
  *b = tmp;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n >> a >> b;
    if (a - b > 1 || b - a > 1 || a + b > n - 2)
      cout << -1 << "\n";
    else {
      int arr[n];
      for (int i = 0; i < n; ++i) arr[i] = i + 1;
      if (b > a) {
        swap(arr[0], arr[1]);
        int rem = min(a, b);
        for (int i = 2; i < n - 2 && rem; i += 2) {
          swap(arr[i], arr[i + 1]);
          rem--;
        }
      } else if (a > b) {
        swap(arr[n - 1], arr[n - 2]);
        int rem = min(a, b);
        for (int i = n - 3; i > 1 && rem; i -= 2) {
          swap(arr[i], arr[i - 1]);
          rem--;
        }
      } else {
        int rem = min(a, b);
        for (int i = 1; i < n - 2 && rem; i += 2) {
          swap(arr[i], arr[i + 1]);
          rem--;
        }
      }
      for (auto x : arr) cout << x << " ";
      cout << "\n";
    }
  }
  return 0;
}

### 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;
void swap(int *a, int *b) {
  int tmp = *a;
  *a = *b;
  *b = tmp;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n >> a >> b;
    if (a - b > 1 || b - a > 1 || a + b > n - 2)
      cout << -1 << "\n";
    else {
      int arr[n];
      for (int i = 0; i < n; ++i) arr[i] = i + 1;
      if (b > a) {
        swap(arr[0], arr[1]);
        int rem = min(a, b);
        for (int i = 2; i < n - 2 && rem; i += 2) {
          swap(arr[i], arr[i + 1]);
          rem--;
        }
      } else if (a > b) {
        swap(arr[n - 1], arr[n - 2]);
        int rem = min(a, b);
        for (int i = n - 3; i > 1 && rem; i -= 2) {
          swap(arr[i], arr[i - 1]);
          rem--;
        }
      } else {
        int rem = min(a, b);
        for (int i = 1; i < n - 2 && rem; i += 2) {
          swap(arr[i], arr[i + 1]);
          rem--;
        }
      }
      for (auto x : arr) cout << x << " ";
      cout << "\n";
    }
  }
  return 0;
}

```