#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long inf = 1000000000000000000;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, a, b;
    cin >> n >> a >> b;
    vector<long long> ans;
    vector<long long> p;
    for (long long i = 1; i < n + 1; i += 1) p.push_back(i);
    if (abs(a - b) > 1 || (a + b) > (n - 2))
      cout << "-1"
           << "\n";
    else {
      long long x = 0, y = 0;
      if (a > b) {
        while (a != 0) {
          ans.push_back(p[0 + y]);
          ans.push_back(p[n - 1 - x]);
          y++;
          x++;
          a--;
          b--;
        }
        for (long long i = n - 1 - x; i >= y; i--) ans.push_back(p[i]);
      } else if (a < b) {
        while (b != 0) {
          ans.push_back(p[n - 1 - x]);
          ans.push_back(p[0 + y]);
          y++;
          x++;
          a--;
          b--;
        }
        for (long long i = y; i <= n - 1 - x; i++) ans.push_back(p[i]);
      } else {
        ans.push_back(1);
        y++;
        while (a != 0) {
          ans.push_back(p[n - 1 - x]);
          ans.push_back(p[0 + y]);
          y++;
          x++;
          a--;
          b--;
        }
        for (long long i = y; i <= n - 1 - x; i++) ans.push_back(p[i]);
      }
      for (long long i = 0; i < n; i += 1) cout << ans[i] << " ";
      cout << "\n";
    }
  }
}

### 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;
const long long mod = 1000000007;
const long long inf = 1000000000000000000;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, a, b;
    cin >> n >> a >> b;
    vector<long long> ans;
    vector<long long> p;
    for (long long i = 1; i < n + 1; i += 1) p.push_back(i);
    if (abs(a - b) > 1 || (a + b) > (n - 2))
      cout << "-1"
           << "\n";
    else {
      long long x = 0, y = 0;
      if (a > b) {
        while (a != 0) {
          ans.push_back(p[0 + y]);
          ans.push_back(p[n - 1 - x]);
          y++;
          x++;
          a--;
          b--;
        }
        for (long long i = n - 1 - x; i >= y; i--) ans.push_back(p[i]);
      } else if (a < b) {
        while (b != 0) {
          ans.push_back(p[n - 1 - x]);
          ans.push_back(p[0 + y]);
          y++;
          x++;
          a--;
          b--;
        }
        for (long long i = y; i <= n - 1 - x; i++) ans.push_back(p[i]);
      } else {
        ans.push_back(1);
        y++;
        while (a != 0) {
          ans.push_back(p[n - 1 - x]);
          ans.push_back(p[0 + y]);
          y++;
          x++;
          a--;
          b--;
        }
        for (long long i = y; i <= n - 1 - x; i++) ans.push_back(p[i]);
      }
      for (long long i = 0; i < n; i += 1) cout << ans[i] << " ";
      cout << "\n";
    }
  }
}

```