#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int arr[N];
int main(void) {
  int T;
  cin >> T;
  while (T--) {
    int n, a, b;
    cin >> n >> a >> b;
    bool flag = true;
    for (int i = 1; i <= n; i++) arr[i] = i;
    int maxv = (n - 2) / 2 + (n & 1);
    if (a > maxv || b > maxv)
      flag = false;
    else if (abs(a - b) > 1)
      flag = false;
    else if ((n & 1) && a == maxv && b == maxv)
      flag = false;
    else if (a == 0 && b == 0) {
      for (int i = 1; i <= n; i++) arr[i] = i;
    } else {
      if (a == b) {
        int l = 1, r = n;
        for (int i = 1; i <= b * 2 + 1; i += 2) arr[i] = r--;
        for (int i = 2; i <= a * 2; i += 2) arr[i] = l++;
        for (int i = 2 * b + 2; i <= n; i++) arr[i] = r--;
      } else if (a == b + 1) {
        int l = 1, r = n;
        for (int i = 1; i <= a * 2; i += 2) arr[i] = l++;
        for (int i = 2; i <= a * 2; i += 2) arr[i] = r--;
        for (int i = 2 * a + 1; i <= n; i++) arr[i] = r--;
      } else {
        int l = 1, r = n;
        for (int i = 1; i <= b * 2; i += 2) arr[i] = r--;
        for (int i = 2; i <= b * 2; i += 2) arr[i] = l++;
        for (int i = 2 * b + 1; i <= n; i++) arr[i] = l++;
      }
    }
    if (flag) {
      for (int i = 1; i <= n; i++) cout << arr[i] << " ";
      cout << endl;
    } else
      cout << "-1" << endl;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 N = 1e5 + 10;
int arr[N];
int main(void) {
  int T;
  cin >> T;
  while (T--) {
    int n, a, b;
    cin >> n >> a >> b;
    bool flag = true;
    for (int i = 1; i <= n; i++) arr[i] = i;
    int maxv = (n - 2) / 2 + (n & 1);
    if (a > maxv || b > maxv)
      flag = false;
    else if (abs(a - b) > 1)
      flag = false;
    else if ((n & 1) && a == maxv && b == maxv)
      flag = false;
    else if (a == 0 && b == 0) {
      for (int i = 1; i <= n; i++) arr[i] = i;
    } else {
      if (a == b) {
        int l = 1, r = n;
        for (int i = 1; i <= b * 2 + 1; i += 2) arr[i] = r--;
        for (int i = 2; i <= a * 2; i += 2) arr[i] = l++;
        for (int i = 2 * b + 2; i <= n; i++) arr[i] = r--;
      } else if (a == b + 1) {
        int l = 1, r = n;
        for (int i = 1; i <= a * 2; i += 2) arr[i] = l++;
        for (int i = 2; i <= a * 2; i += 2) arr[i] = r--;
        for (int i = 2 * a + 1; i <= n; i++) arr[i] = r--;
      } else {
        int l = 1, r = n;
        for (int i = 1; i <= b * 2; i += 2) arr[i] = r--;
        for (int i = 2; i <= b * 2; i += 2) arr[i] = l++;
        for (int i = 2 * b + 1; i <= n; i++) arr[i] = l++;
      }
    }
    if (flag) {
      for (int i = 1; i <= n; i++) cout << arr[i] << " ";
      cout << endl;
    } else
      cout << "-1" << endl;
  }
  return 0;
}

```