#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long infi = INT_MAX;
const long long infll = LLONG_MAX;
const long double PI = 3.1415926535897932384626;
void solve(long long test_case) {
  long long n, a, b;
  cin >> n >> a >> b;
  if (a + b + 2 > n) {
    cout << -1 << "\n";
    return;
  }
  if (abs(a - b) > 1) {
    cout << -1 << "\n";
    return;
  }
  vector<long long> arr(n);
  for (long long i = 0; i < n; i++) arr[i] = i + 1;
  if (a > b) {
    for (long long i = n - 1; i - 1 >= 0; i -= 2) {
      swap(arr[i], arr[i - 1]);
      if (--a == 0) break;
    }
  } else if (b > a) {
    for (long long i = 0; i + 1 < n; i += 2) {
      swap(arr[i], arr[i + 1]);
      if (--b == 0) break;
    }
  } else {
    for (long long i = 1; i + 1 < n; i += 2) {
      if (a == 0) break;
      swap(arr[i], arr[i + 1]);
      a--;
    }
  }
  for (auto x : arr) cout << x << " ";
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  cin >> tc;
  while (tc--) solve(tc);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 long long MOD = 1e9 + 7;
const long long infi = INT_MAX;
const long long infll = LLONG_MAX;
const long double PI = 3.1415926535897932384626;
void solve(long long test_case) {
  long long n, a, b;
  cin >> n >> a >> b;
  if (a + b + 2 > n) {
    cout << -1 << "\n";
    return;
  }
  if (abs(a - b) > 1) {
    cout << -1 << "\n";
    return;
  }
  vector<long long> arr(n);
  for (long long i = 0; i < n; i++) arr[i] = i + 1;
  if (a > b) {
    for (long long i = n - 1; i - 1 >= 0; i -= 2) {
      swap(arr[i], arr[i - 1]);
      if (--a == 0) break;
    }
  } else if (b > a) {
    for (long long i = 0; i + 1 < n; i += 2) {
      swap(arr[i], arr[i + 1]);
      if (--b == 0) break;
    }
  } else {
    for (long long i = 1; i + 1 < n; i += 2) {
      if (a == 0) break;
      swap(arr[i], arr[i + 1]);
      a--;
    }
  }
  for (auto x : arr) cout << x << " ";
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  cin >> tc;
  while (tc--) solve(tc);
  return 0;
}

```