#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n >> a >> b;
    vector<int> v(n, 0);
    if (a + b <= n - 2 && abs(a - b) <= 1) {
      if (a > b) {
        for (int i = 1, j = 1; j <= a; j++, i += 2) v[i] = n - a + j;
        for (int i = n - 1, j = 1; i >= 0; i--) {
          if (!v[i]) {
            v[i] = j;
            j++;
          }
        }
      } else if (b > a) {
        for (int i = 1, j = b; j >= 1; j--, i += 2) v[i] = j;
        for (int i = n - 1, j = n; i >= 0; i--) {
          if (!v[i]) {
            v[i] = j;
            j--;
          }
        }
      } else {
        for (int i = 1, j = 1; j <= a; j++, i += 2) v[i] = n - a + j;
        for (int i = 0, j = b; j >= 1; j--, i += 2) v[i] = j;
        for (int i = n - 1, j = n - a; i >= 0; i--) {
          if (!v[i]) {
            v[i] = j;
            j--;
          }
        }
      }
      for (int i = 0; i < n; i++) cout << v[i] << ' ';
      cout << '\n';
    } else
      cout << "-1\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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n >> a >> b;
    vector<int> v(n, 0);
    if (a + b <= n - 2 && abs(a - b) <= 1) {
      if (a > b) {
        for (int i = 1, j = 1; j <= a; j++, i += 2) v[i] = n - a + j;
        for (int i = n - 1, j = 1; i >= 0; i--) {
          if (!v[i]) {
            v[i] = j;
            j++;
          }
        }
      } else if (b > a) {
        for (int i = 1, j = b; j >= 1; j--, i += 2) v[i] = j;
        for (int i = n - 1, j = n; i >= 0; i--) {
          if (!v[i]) {
            v[i] = j;
            j--;
          }
        }
      } else {
        for (int i = 1, j = 1; j <= a; j++, i += 2) v[i] = n - a + j;
        for (int i = 0, j = b; j >= 1; j--, i += 2) v[i] = j;
        for (int i = n - 1, j = n - a; i >= 0; i--) {
          if (!v[i]) {
            v[i] = j;
            j--;
          }
        }
      }
      for (int i = 0; i < n; i++) cout << v[i] << ' ';
      cout << '\n';
    } else
      cout << "-1\n";
  }
}

```