#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
void solve() {
  long long n, a, b;
  cin >> n >> a >> b;
  if (a + b > n - 2) {
    cout << -1;
    cout << "\n";
    return;
  }
  if (n % 2 == 0 && (a > (n - 2) / 2 || b > (n - 2) / 2)) {
    cout << -1;
    cout << "\n";
    return;
  }
  if (n % 2 && (a > (n - 1) / 2 || b > (n - 1) / 2)) {
    cout << -1;
    cout << "\n";
    return;
  }
  if (abs(a - b) > 1) {
    cout << -1;
    cout << "\n";
    return;
  }
  if (a == b) {
    cout << 1 << " ";
    long long x = 2;
    long long y = n;
    while (a--) {
      cout << y << " " << x << " ";
      x++;
      y--;
    }
    for (long long i = x; i <= y; i++) cout << i << " ";
    cout << "\n";
    return;
  }
  if (a > b) {
    cout << 1 << " " << n << " ";
    long long x = 2;
    long long y = n - 1;
    while (b--) {
      cout << x << " " << y << " ";
      x++;
      y--;
    }
    for (long long i = y; i >= x; i--) cout << i << " ";
    cout << "\n";
    return;
  }
  cout << n << " " << 1 << " ";
  long long x = n - 1;
  long long y = 2;
  while (a--) {
    cout << x << " " << y << " ";
    x--;
    y++;
  }
  for (long long i = y; i <= x; i++) cout << i << " ";
  cout << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

### 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;
const long long mod = 1000000007;
void solve() {
  long long n, a, b;
  cin >> n >> a >> b;
  if (a + b > n - 2) {
    cout << -1;
    cout << "\n";
    return;
  }
  if (n % 2 == 0 && (a > (n - 2) / 2 || b > (n - 2) / 2)) {
    cout << -1;
    cout << "\n";
    return;
  }
  if (n % 2 && (a > (n - 1) / 2 || b > (n - 1) / 2)) {
    cout << -1;
    cout << "\n";
    return;
  }
  if (abs(a - b) > 1) {
    cout << -1;
    cout << "\n";
    return;
  }
  if (a == b) {
    cout << 1 << " ";
    long long x = 2;
    long long y = n;
    while (a--) {
      cout << y << " " << x << " ";
      x++;
      y--;
    }
    for (long long i = x; i <= y; i++) cout << i << " ";
    cout << "\n";
    return;
  }
  if (a > b) {
    cout << 1 << " " << n << " ";
    long long x = 2;
    long long y = n - 1;
    while (b--) {
      cout << x << " " << y << " ";
      x++;
      y--;
    }
    for (long long i = y; i >= x; i--) cout << i << " ";
    cout << "\n";
    return;
  }
  cout << n << " " << 1 << " ";
  long long x = n - 1;
  long long y = 2;
  while (a--) {
    cout << x << " " << y << " ";
    x--;
    y++;
  }
  for (long long i = y; i <= x; i++) cout << i << " ";
  cout << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

```