#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int num[100001];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int t, n, a, b;
  cin >> t;
  while (t--) {
    cin >> n >> a >> b;
    if (abs(a - b) > 1 || a + b > n - 2) {
      cout << -1 << endl;
      continue;
    }
    if (a > b)
      for (int i = 1; i <= n; i++) {
        num[i] = n - i + 1;
      }
    else if (a < b) {
      for (int i = 1; i <= n; i++) {
        num[i] = i;
      }
    } else {
      for (int i = 1; i <= n; i++) {
        num[i] = i;
      }
      for (int i = 1; i <= a; i++) swap(num[i * 2], num[i * 2 + 1]);
      for (int i = 1; i <= n; i++) cout << num[i] << ' ';
      cout << endl;
      continue;
    }
    int cnt = max(a, b);
    for (int i = 1; i <= cnt; i++) {
      swap(num[i * 2 - 1], num[i * 2]);
    }
    for (int i = 1; i <= n; i++) cout << num[i] << " ";
    cout << endl;
  }
  return 0;
}

### 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;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int num[100001];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int t, n, a, b;
  cin >> t;
  while (t--) {
    cin >> n >> a >> b;
    if (abs(a - b) > 1 || a + b > n - 2) {
      cout << -1 << endl;
      continue;
    }
    if (a > b)
      for (int i = 1; i <= n; i++) {
        num[i] = n - i + 1;
      }
    else if (a < b) {
      for (int i = 1; i <= n; i++) {
        num[i] = i;
      }
    } else {
      for (int i = 1; i <= n; i++) {
        num[i] = i;
      }
      for (int i = 1; i <= a; i++) swap(num[i * 2], num[i * 2 + 1]);
      for (int i = 1; i <= n; i++) cout << num[i] << ' ';
      cout << endl;
      continue;
    }
    int cnt = max(a, b);
    for (int i = 1; i <= cnt; i++) {
      swap(num[i * 2 - 1], num[i * 2]);
    }
    for (int i = 1; i <= n; i++) cout << num[i] << " ";
    cout << endl;
  }
  return 0;
}

```