#include <bits/stdc++.h>
using namespace std;
constexpr int N = 3e5 + 10;
int arr[N];
void solve() {
  int n, a, b;
  cin >> n >> a >> b;
  if (a + b + 2 > n) {
    cout << -1 << endl;
    return;
  }
  if (abs(a - b) > 1) {
    cout << -1 << endl;
    return;
  }
  if (a == b) {
    for (int i = a + b + 2; i < n; i++) {
      arr[i] = i + 1;
    }
    for (int i = 0; i <= b; i++) {
      arr[i * 2 + 1] = a + i + 2;
    }
    for (int i = 0; i <= a; i++) {
      arr[i * 2] = i + 1;
    }
  } else if (a < b) {
    for (int i = 0; i <= b; i++) {
      arr[i * 2] = b + i + 1;
    }
    for (int i = 0; i <= a; i++) {
      arr[i * 2 + 1] = i + 1;
    }
    for (int i = a + b + 2; i < n; i++) {
      arr[i] = i + 1;
    }
  } else {
    int cnt = n - (a + b + 2);
    for (int i = 0; i <= a; i++) {
      arr[i * 2] = cnt + i + 1;
    }
    for (int i = 0; i <= b; i++) {
      arr[i * 2 + 1] = n - i;
    }
    for (int i = n - cnt; i < n; i++) {
      arr[i] = n - i;
    }
  }
  if (a == 0 && b == 0) {
    for (int i = 0; i < n; i++) {
      arr[i] = i + 1;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << arr[i] << ' ';
  }
  cout << endl;
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
  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;
constexpr int N = 3e5 + 10;
int arr[N];
void solve() {
  int n, a, b;
  cin >> n >> a >> b;
  if (a + b + 2 > n) {
    cout << -1 << endl;
    return;
  }
  if (abs(a - b) > 1) {
    cout << -1 << endl;
    return;
  }
  if (a == b) {
    for (int i = a + b + 2; i < n; i++) {
      arr[i] = i + 1;
    }
    for (int i = 0; i <= b; i++) {
      arr[i * 2 + 1] = a + i + 2;
    }
    for (int i = 0; i <= a; i++) {
      arr[i * 2] = i + 1;
    }
  } else if (a < b) {
    for (int i = 0; i <= b; i++) {
      arr[i * 2] = b + i + 1;
    }
    for (int i = 0; i <= a; i++) {
      arr[i * 2 + 1] = i + 1;
    }
    for (int i = a + b + 2; i < n; i++) {
      arr[i] = i + 1;
    }
  } else {
    int cnt = n - (a + b + 2);
    for (int i = 0; i <= a; i++) {
      arr[i * 2] = cnt + i + 1;
    }
    for (int i = 0; i <= b; i++) {
      arr[i * 2 + 1] = n - i;
    }
    for (int i = n - cnt; i < n; i++) {
      arr[i] = n - i;
    }
  }
  if (a == 0 && b == 0) {
    for (int i = 0; i < n; i++) {
      arr[i] = i + 1;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << arr[i] << ' ';
  }
  cout << endl;
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

```