#include <bits/stdc++.h>
using namespace std;
bool fun(int arr[], int n, int a, int b) {
  if (abs(a - b) > 1) return false;
  int j = 1;
  for (int i = 0; i < n; i++) {
    arr[i] = j;
    j++;
  }
  if (a == b) {
    if ((2 * a + 1) + 1 > n) return false;
    int ct = 1;
    for (int i = 0; i < a + 1; i++) {
      arr[2 * i] = ct;
      ct++;
    }
    for (int i = 0; i < a + 1; i++) {
      arr[2 * i + 1] = ct;
      ct++;
    }
    return true;
  }
  if (b > a) {
    int ct = 1;
    if (2 * b + 1 > n) return false;
    for (int i = 0; i < b; i++) {
      arr[2 * i + 1] = ct;
      ct++;
    }
    for (int i = 0; i < b + 1; i++) {
      arr[2 * i] = ct;
      ct++;
    }
    return true;
  }
  if (a > b) {
    int ct = n - 2 * a;
    if (2 * a + 1 > n) return false;
    for (int i = 0; i < a + 1; i++) {
      arr[2 * i + n - 2 * a - 1] = ct;
      ct++;
    }
    for (int i = 0; i < a; i++) {
      arr[2 * i + 1 + n - 2 * a - 1] = ct;
      ct++;
    }
    return true;
  }
  return false;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n >> a >> b;
    int arr[n];
    if (!fun(arr, n, a, b))
      cout << -1 << "\n";
    else {
      for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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;
bool fun(int arr[], int n, int a, int b) {
  if (abs(a - b) > 1) return false;
  int j = 1;
  for (int i = 0; i < n; i++) {
    arr[i] = j;
    j++;
  }
  if (a == b) {
    if ((2 * a + 1) + 1 > n) return false;
    int ct = 1;
    for (int i = 0; i < a + 1; i++) {
      arr[2 * i] = ct;
      ct++;
    }
    for (int i = 0; i < a + 1; i++) {
      arr[2 * i + 1] = ct;
      ct++;
    }
    return true;
  }
  if (b > a) {
    int ct = 1;
    if (2 * b + 1 > n) return false;
    for (int i = 0; i < b; i++) {
      arr[2 * i + 1] = ct;
      ct++;
    }
    for (int i = 0; i < b + 1; i++) {
      arr[2 * i] = ct;
      ct++;
    }
    return true;
  }
  if (a > b) {
    int ct = n - 2 * a;
    if (2 * a + 1 > n) return false;
    for (int i = 0; i < a + 1; i++) {
      arr[2 * i + n - 2 * a - 1] = ct;
      ct++;
    }
    for (int i = 0; i < a; i++) {
      arr[2 * i + 1 + n - 2 * a - 1] = ct;
      ct++;
    }
    return true;
  }
  return false;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n >> a >> b;
    int arr[n];
    if (!fun(arr, n, a, b))
      cout << -1 << "\n";
    else {
      for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}

```