#include <bits/stdc++.h>
using namespace std;
int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n >> a >> b;
    if (abs(a - b) > 1) {
      cout << -1 << '\n';
      continue;
    } else if (a + b + 2 > n) {
      cout << -1 << '\n';
      continue;
    }
    int o = 3, e = 2;
    if (a == b) {
      for (long long int i = 1; i < 2 * a + 2; i++) {
        if (i == 1)
          cout << 1 << " ";
        else if (i & 1) {
          cout << e << " ";
          e += 2;
        } else {
          cout << o << " ";
          o += 2;
        }
      }
      for (long long int i = 2 * a + 2; i < n + 1; i++) cout << i << " ";
      cout << '\n';
    } else if (a > b) {
      for (long long int i = 1; i < 2 * b + 2; i++) {
        if (i == 1)
          cout << 1 << " ";
        else if (i & 1) {
          cout << e << " ";
          e += 2;
        } else {
          cout << o << " ";
          o += 2;
        }
      }
      for (long long int i = n + 1 - 1; i >= 2 * b + 2; i--) cout << i << " ";
      cout << '\n';
    } else if (a < b) {
      o = b + 1;
      e = 1;
      for (long long int i = 1; i < 2 * b + 1; i++) {
        if (i & 1) {
          cout << o << " ";
          o += 1;
        } else {
          cout << e << " ";
          e += 1;
        }
      }
      for (long long int i = 2 * b + 1; i < n + 1; i++) cout << i << " ";
      cout << '\n';
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n >> a >> b;
    if (abs(a - b) > 1) {
      cout << -1 << '\n';
      continue;
    } else if (a + b + 2 > n) {
      cout << -1 << '\n';
      continue;
    }
    int o = 3, e = 2;
    if (a == b) {
      for (long long int i = 1; i < 2 * a + 2; i++) {
        if (i == 1)
          cout << 1 << " ";
        else if (i & 1) {
          cout << e << " ";
          e += 2;
        } else {
          cout << o << " ";
          o += 2;
        }
      }
      for (long long int i = 2 * a + 2; i < n + 1; i++) cout << i << " ";
      cout << '\n';
    } else if (a > b) {
      for (long long int i = 1; i < 2 * b + 2; i++) {
        if (i == 1)
          cout << 1 << " ";
        else if (i & 1) {
          cout << e << " ";
          e += 2;
        } else {
          cout << o << " ";
          o += 2;
        }
      }
      for (long long int i = n + 1 - 1; i >= 2 * b + 2; i--) cout << i << " ";
      cout << '\n';
    } else if (a < b) {
      o = b + 1;
      e = 1;
      for (long long int i = 1; i < 2 * b + 1; i++) {
        if (i & 1) {
          cout << o << " ";
          o += 1;
        } else {
          cout << e << " ";
          e += 1;
        }
      }
      for (long long int i = 2 * b + 1; i < n + 1; i++) cout << i << " ";
      cout << '\n';
    }
  }
  return 0;
}

```