#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long max_n = 100005;
long long arr[max_n];
void solve() {
  long long n, a, b;
  cin >> n >> a >> b;
  if (a + b > n - 2) {
    cout << -1 << "\n";
    return;
  }
  if (a == b + 1) {
    for (long long i = 1; i <= b + 1; i++) {
      arr[i * 2 - 1] = i + n - (2 * b + 2);
      arr[i * 2] = arr[i * 2 - 1] + (b + 1);
    }
    for (long long i = 2 * (b + 1) + 1; i <= n; i++) arr[i] = (n + 1) - i;
  } else if (a == b) {
    for (long long i = 1; i <= b; i++) {
      arr[i * 2 - 1] = i + b + n - (2 * b + 1);
      arr[i * 2] = i + n - (2 * b + 1);
    }
    arr[2 * b + 1] = n;
    for (long long i = 2 * b + 2; i <= n; i++) arr[i] = (n + 1) - i;
  } else if (a + 1 == b) {
    swap(a, b);
    for (long long i = 1; i <= b + 1; i++) {
      arr[i * 2 - 1] = i + n - (2 * b + 2);
      arr[i * 2] = arr[i * 2 - 1] + (b + 1);
    }
    for (long long i = 2 * (b + 1) + 1; i <= n; i++) arr[i] = (n + 1) - i;
    for (long long i = 1; i <= n; i++) arr[i] = (n + 1) - arr[i];
    swap(a, b);
  } else {
    cout << -1 << "\n";
    return;
  }
  long long cnt_a = 0, cnt_b = 0;
  for (long long i = 2; i < n; i++) {
    if ((arr[i] < arr[i - 1]) && (arr[i] < arr[i + 1]))
      cnt_b++;
    else if ((arr[i] > arr[i - 1]) && (arr[i] > arr[i + 1]))
      cnt_a++;
  }
  assert(cnt_a == a);
  assert(cnt_b == b);
  for (long long i = 1; i <= n; i++) cout << arr[i] << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
}

### 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>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long max_n = 100005;
long long arr[max_n];
void solve() {
  long long n, a, b;
  cin >> n >> a >> b;
  if (a + b > n - 2) {
    cout << -1 << "\n";
    return;
  }
  if (a == b + 1) {
    for (long long i = 1; i <= b + 1; i++) {
      arr[i * 2 - 1] = i + n - (2 * b + 2);
      arr[i * 2] = arr[i * 2 - 1] + (b + 1);
    }
    for (long long i = 2 * (b + 1) + 1; i <= n; i++) arr[i] = (n + 1) - i;
  } else if (a == b) {
    for (long long i = 1; i <= b; i++) {
      arr[i * 2 - 1] = i + b + n - (2 * b + 1);
      arr[i * 2] = i + n - (2 * b + 1);
    }
    arr[2 * b + 1] = n;
    for (long long i = 2 * b + 2; i <= n; i++) arr[i] = (n + 1) - i;
  } else if (a + 1 == b) {
    swap(a, b);
    for (long long i = 1; i <= b + 1; i++) {
      arr[i * 2 - 1] = i + n - (2 * b + 2);
      arr[i * 2] = arr[i * 2 - 1] + (b + 1);
    }
    for (long long i = 2 * (b + 1) + 1; i <= n; i++) arr[i] = (n + 1) - i;
    for (long long i = 1; i <= n; i++) arr[i] = (n + 1) - arr[i];
    swap(a, b);
  } else {
    cout << -1 << "\n";
    return;
  }
  long long cnt_a = 0, cnt_b = 0;
  for (long long i = 2; i < n; i++) {
    if ((arr[i] < arr[i - 1]) && (arr[i] < arr[i + 1]))
      cnt_b++;
    else if ((arr[i] > arr[i - 1]) && (arr[i] > arr[i + 1]))
      cnt_a++;
  }
  assert(cnt_a == a);
  assert(cnt_b == b);
  for (long long i = 1; i <= n; i++) cout << arr[i] << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
}

```