#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n = 10, a, b;
    cin >> n >> a >> b;
    if (abs(a - b) > 1 || (a + b + 2) > n)
      cout << -1 << '\n';
    else {
      vector<int> ans(n);
      iota(ans.begin(), ans.end(), 1);
      int i = a >= b ? n - 1 : 0;
      int j = a >= b ? -2 : 2;
      int k = a >= b ? -1 : 1;
      int x = max(a, b);
      if (a == b) i--;
      while (x--) {
        swap(ans[i], ans[i + k]);
        i += j;
      }
      int g = 0, d = 0;
      for (int i = 1; i < n - 1; ++i) {
        if (ans[i - 1] > ans[i] && ans[i] < ans[i + 1]) g++;
        if (ans[i - 1] < ans[i] && ans[i + 1] < ans[i]) d++;
      }
      for (auto &i : ans) {
        cout << i << " ";
      }
      cout << '\n';
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n = 10, a, b;
    cin >> n >> a >> b;
    if (abs(a - b) > 1 || (a + b + 2) > n)
      cout << -1 << '\n';
    else {
      vector<int> ans(n);
      iota(ans.begin(), ans.end(), 1);
      int i = a >= b ? n - 1 : 0;
      int j = a >= b ? -2 : 2;
      int k = a >= b ? -1 : 1;
      int x = max(a, b);
      if (a == b) i--;
      while (x--) {
        swap(ans[i], ans[i + k]);
        i += j;
      }
      int g = 0, d = 0;
      for (int i = 1; i < n - 1; ++i) {
        if (ans[i - 1] > ans[i] && ans[i] < ans[i + 1]) g++;
        if (ans[i - 1] < ans[i] && ans[i + 1] < ans[i]) d++;
      }
      for (auto &i : ans) {
        cout << i << " ";
      }
      cout << '\n';
    }
  }
  return 0;
}

```