#include <bits/stdc++.h>
using namespace std;
const int Mod = 1e7;
const int N = 1e5 + 10;
long long T = 1;
long long s[N];
void init(long long n) {
  for (long long i = 0; i <= n; i++) s[i] = i;
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin >> T;
  while (T--) {
    long long n, a, b;
    std::ios::sync_with_stdio(false);
    cin >> n >> a >> b;
    init(n);
    long long mid = (n / 2) - 1;
    if (n % 2 == 1) mid++;
    long long x = abs(a - b);
    if (x <= 1 && (a + b) <= (n - 2) && a <= mid && b <= mid) {
      if (a == b) {
        for (long long i = 2, j = a; j > 0; i += 2, j--) {
          swap(s[i], s[i + 1]);
        }
      }
      if (a > b) {
        swap(s[n], s[n - 1]);
        for (long long i = 2, j = b; j > 0; j--, i += 2) swap(s[i], s[i + 1]);
      }
      if (b > a) {
        swap(s[1], s[2]);
        for (long long i = n - 1, j = a; j > 0; j--, i -= 2)
          swap(s[i], s[i - 1]);
      }
      for (long long i = 1; i <= n; i++) cout << s[i] << " ";
      cout << '\n';
    } else {
      cout << "-1" << '\n';
    }
  }
  return 0;
}

### Prompt

Generate a cpp solution 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;
const int Mod = 1e7;
const int N = 1e5 + 10;
long long T = 1;
long long s[N];
void init(long long n) {
  for (long long i = 0; i <= n; i++) s[i] = i;
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin >> T;
  while (T--) {
    long long n, a, b;
    std::ios::sync_with_stdio(false);
    cin >> n >> a >> b;
    init(n);
    long long mid = (n / 2) - 1;
    if (n % 2 == 1) mid++;
    long long x = abs(a - b);
    if (x <= 1 && (a + b) <= (n - 2) && a <= mid && b <= mid) {
      if (a == b) {
        for (long long i = 2, j = a; j > 0; i += 2, j--) {
          swap(s[i], s[i + 1]);
        }
      }
      if (a > b) {
        swap(s[n], s[n - 1]);
        for (long long i = 2, j = b; j > 0; j--, i += 2) swap(s[i], s[i + 1]);
      }
      if (b > a) {
        swap(s[1], s[2]);
        for (long long i = n - 1, j = a; j > 0; j--, i -= 2)
          swap(s[i], s[i - 1]);
      }
      for (long long i = 1; i <= n; i++) cout << s[i] << " ";
      cout << '\n';
    } else {
      cout << "-1" << '\n';
    }
  }
  return 0;
}

```