#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, a, b;
  cin >> n >> a >> b;
  if (abs(a - b) > 1) {
    cout << "-1\n";
    return;
  }
  if (a > b) {
    if (2 * a + 1 > n) {
      cout << "-1\n";
      return;
    }
    cout << "1 ";
    for (long long i = a + 2; i <= n - a; i++) cout << i << " ";
    for (long long i = 2; i <= a + 1; i++) {
      cout << n - i + 2 << " " << i << " ";
    }
    cout << "\n";
  } else if (b > a) {
    if (2 * b + 1 > n) {
      cout << "-1\n";
      return;
    }
    cout << n << " ";
    for (long long i = n - b - 1; i > b; i--) {
      cout << i << " ";
    }
    for (long long i = 1; i <= b; i++) {
      cout << i << " " << n - i << " ";
    }
    cout << "\n";
  } else {
    if (2 * a + 2 > n) {
      cout << "-1\n";
      return;
    }
    if (n == 1) {
      cout << "1\n";
      return;
    }
    cout << "1 ";
    for (long long i = a + 2; i < n - a; i++) cout << i << " ";
    cout << n << " ";
    for (long long i = 2; i <= a + 1; i++) {
      cout << i << " " << n - i + 1 << " ";
    }
    cout << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long testcases = 1;
  cin >> testcases;
  while (testcases--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
void solve() {
  long long n, a, b;
  cin >> n >> a >> b;
  if (abs(a - b) > 1) {
    cout << "-1\n";
    return;
  }
  if (a > b) {
    if (2 * a + 1 > n) {
      cout << "-1\n";
      return;
    }
    cout << "1 ";
    for (long long i = a + 2; i <= n - a; i++) cout << i << " ";
    for (long long i = 2; i <= a + 1; i++) {
      cout << n - i + 2 << " " << i << " ";
    }
    cout << "\n";
  } else if (b > a) {
    if (2 * b + 1 > n) {
      cout << "-1\n";
      return;
    }
    cout << n << " ";
    for (long long i = n - b - 1; i > b; i--) {
      cout << i << " ";
    }
    for (long long i = 1; i <= b; i++) {
      cout << i << " " << n - i << " ";
    }
    cout << "\n";
  } else {
    if (2 * a + 2 > n) {
      cout << "-1\n";
      return;
    }
    if (n == 1) {
      cout << "1\n";
      return;
    }
    cout << "1 ";
    for (long long i = a + 2; i < n - a; i++) cout << i << " ";
    cout << n << " ";
    for (long long i = 2; i <= a + 1; i++) {
      cout << i << " " << n - i + 1 << " ";
    }
    cout << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long testcases = 1;
  cin >> testcases;
  while (testcases--) solve();
  return 0;
}

```