#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int n, m, t, q, a, b, c, h, d, x, y, z, i, j, k,
      ans = 0, ans2 = 0, cur = 0, cur2 = 0, sum = 0, sum2 = 0, maxn;
  string str = "", str2, str3;
  cin >> t;
  while (t--) {
    cin >> n >> x >> y;
    if (abs(x - y) > 1) {
      cout << "-1\n";
      continue;
    } else if (x + y > n - 2) {
      cout << "-1\n";
      continue;
    }
    a = 1;
    b = n;
    if (x == y) {
      cout << "1 ";
      a++;
      for (i = 0; i < x; i++) {
        cout << b << " " << a << " ";
        a++;
        b--;
      }
      for (i = a; i <= b; i++) cout << i << " ";
      cout << "\n";
    } else if (x > y) {
      for (i = 0; i < x; i++) {
        cout << a << " " << b << " ";
        a++;
        b--;
      }
      for (i = b; i >= a; i--) cout << i << " ";
      cout << "\n";
    } else {
      for (i = 0; i < y; i++) {
        cout << b << " " << a << " ";
        a++;
        b--;
      }
      for (i = a; i <= b; i++) cout << i << " ";
      cout << "\n";
    }
  }
}

### Prompt

Please create a solution in CPP 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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int n, m, t, q, a, b, c, h, d, x, y, z, i, j, k,
      ans = 0, ans2 = 0, cur = 0, cur2 = 0, sum = 0, sum2 = 0, maxn;
  string str = "", str2, str3;
  cin >> t;
  while (t--) {
    cin >> n >> x >> y;
    if (abs(x - y) > 1) {
      cout << "-1\n";
      continue;
    } else if (x + y > n - 2) {
      cout << "-1\n";
      continue;
    }
    a = 1;
    b = n;
    if (x == y) {
      cout << "1 ";
      a++;
      for (i = 0; i < x; i++) {
        cout << b << " " << a << " ";
        a++;
        b--;
      }
      for (i = a; i <= b; i++) cout << i << " ";
      cout << "\n";
    } else if (x > y) {
      for (i = 0; i < x; i++) {
        cout << a << " " << b << " ";
        a++;
        b--;
      }
      for (i = b; i >= a; i--) cout << i << " ";
      cout << "\n";
    } else {
      for (i = 0; i < y; i++) {
        cout << b << " " << a << " ";
        a++;
        b--;
      }
      for (i = a; i <= b; i++) cout << i << " ";
      cout << "\n";
    }
  }
}

```