#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, a, b, i, j;
  cin >> n >> a >> b;
  if ((fabs(a - b) == 1 || a == b) && a + b <= n - 2) {
    int ans[n + 3];
    for (i = 1; i <= n; i++) ans[i] = i;
    if (a == b) {
      int x = 2;
      while (a) {
        swap(ans[x], ans[x + 1]);
        x += 2;
        a--;
      }
    } else {
      if (a > b) {
        int x = 2;
        while (b) {
          swap(ans[x], ans[x + 1]);
          x += 2;
          b--;
        }
        swap(ans[n], ans[n - 1]);
      } else {
        int x = n - 1;
        while (a) {
          swap(ans[x], ans[x - 1]);
          x -= 2;
          a--;
        }
        swap(ans[1], ans[2]);
      }
    }
    for (i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << endl;
  } else {
    cout << -1 << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
void solve() {
  int n, a, b, i, j;
  cin >> n >> a >> b;
  if ((fabs(a - b) == 1 || a == b) && a + b <= n - 2) {
    int ans[n + 3];
    for (i = 1; i <= n; i++) ans[i] = i;
    if (a == b) {
      int x = 2;
      while (a) {
        swap(ans[x], ans[x + 1]);
        x += 2;
        a--;
      }
    } else {
      if (a > b) {
        int x = 2;
        while (b) {
          swap(ans[x], ans[x + 1]);
          x += 2;
          b--;
        }
        swap(ans[n], ans[n - 1]);
      } else {
        int x = n - 1;
        while (a) {
          swap(ans[x], ans[x - 1]);
          x -= 2;
          a--;
        }
        swap(ans[1], ans[2]);
      }
    }
    for (i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << endl;
  } else {
    cout << -1 << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```