#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, a, b;
  cin >> n >> a >> b;
  vector<long long> res;
  for (long long i = 1; i <= n; i++) res.push_back(i);
  if ((abs(a - b)) > 1 || ((a + b) > n - 2)) {
    cout << -1 << endl;
    return;
  } else {
    long long i = 1, lim = n;
    if (a > b) {
      res[n - 1] = res[n - 2];
      res[n - 2] = n;
      a--;
      lim = n - 2;
      i = 1;
    }
    if (b > a) {
      res[0] = 2;
      res[1] = 1;
      b--;
      i = 2;
    }
    for (; i < lim && a > 0 && b > 0; i += 2) {
      res[i + 1] = res[i];
      res[i] = i + 2;
      a--, b--;
    }
    for (long long i = 0; i < n; i++) cout << res[i] << " ";
    cout << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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;
  vector<long long> res;
  for (long long i = 1; i <= n; i++) res.push_back(i);
  if ((abs(a - b)) > 1 || ((a + b) > n - 2)) {
    cout << -1 << endl;
    return;
  } else {
    long long i = 1, lim = n;
    if (a > b) {
      res[n - 1] = res[n - 2];
      res[n - 2] = n;
      a--;
      lim = n - 2;
      i = 1;
    }
    if (b > a) {
      res[0] = 2;
      res[1] = 1;
      b--;
      i = 2;
    }
    for (; i < lim && a > 0 && b > 0; i += 2) {
      res[i + 1] = res[i];
      res[i] = i + 2;
      a--, b--;
    }
    for (long long i = 0; i < n; i++) cout << res[i] << " ";
    cout << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```