#include <bits/stdc++.h>
using namespace std;
struct tri {
  long long t1, t2, t3;
};
bool comp(tri a, tri b) { return a.t1 < b.t1; }
void solvee() {
  int n;
  cin >> n;
  int a, b;
  cin >> a >> b;
  if (a == 0 && b == 0) {
    for (int i = 1; i <= n; i++) cout << i << " ";
    cout << "\n";
    return;
  }
  if (abs(a - b) > 1 || a + b + 2 > n) {
    cout << "-1\n";
  } else {
    int t = 0;
    int p[n + 1];
    if (a > b) {
      for (int i = 1, k = 1; k <= a; i += 2, k++) p[i] = k;
      for (int i = 2, k = n; k > n - a; i += 2, k--) p[i] = k;
      for (int i = a * 2 + 1, k = n - a; i <= n; i++, k--) p[i] = k;
    }
    if (a <= b) {
      for (int i = 2, k = 1; k <= b; i += 2, k++) p[i] = k;
      for (int i = 1, k = n; k > n - b; i += 2, k--) p[i] = k;
      if (a == b)
        for (int i = b * 2 + 1, k = n - b; i <= n; k--, i++) p[i] = k;
      else
        for (int i = b * 2 + 1, k = b + 1; i <= n; k++, i++) p[i] = k;
    }
    for (int i = 1; i <= n; i++) cout << p[i] << " ";
    cout << "\n";
    return;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  int tt = 1;
  cin >> tt;
  while (tt--) {
    solvee();
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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;
struct tri {
  long long t1, t2, t3;
};
bool comp(tri a, tri b) { return a.t1 < b.t1; }
void solvee() {
  int n;
  cin >> n;
  int a, b;
  cin >> a >> b;
  if (a == 0 && b == 0) {
    for (int i = 1; i <= n; i++) cout << i << " ";
    cout << "\n";
    return;
  }
  if (abs(a - b) > 1 || a + b + 2 > n) {
    cout << "-1\n";
  } else {
    int t = 0;
    int p[n + 1];
    if (a > b) {
      for (int i = 1, k = 1; k <= a; i += 2, k++) p[i] = k;
      for (int i = 2, k = n; k > n - a; i += 2, k--) p[i] = k;
      for (int i = a * 2 + 1, k = n - a; i <= n; i++, k--) p[i] = k;
    }
    if (a <= b) {
      for (int i = 2, k = 1; k <= b; i += 2, k++) p[i] = k;
      for (int i = 1, k = n; k > n - b; i += 2, k--) p[i] = k;
      if (a == b)
        for (int i = b * 2 + 1, k = n - b; i <= n; k--, i++) p[i] = k;
      else
        for (int i = b * 2 + 1, k = b + 1; i <= n; k++, i++) p[i] = k;
    }
    for (int i = 1; i <= n; i++) cout << p[i] << " ";
    cout << "\n";
    return;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  int tt = 1;
  cin >> tt;
  while (tt--) {
    solvee();
  }
  return 0;
}

```