#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n >> a >> b;
    int N = n;
    if (abs(a - b) > 1) {
      cout << -1 << "\n";
      continue;
    }
    if (a + b > n - 2) {
      cout << -1 << "\n";
      continue;
    }
    bool flag = false;
    if (a < b) {
      swap(a, b);
      flag = true;
    }
    int A = a;
    int B = b;
    int sol[n];
    int i = 1;
    while (a--) {
      sol[i] = n;
      i += 2;
      n--;
    }
    i = 2;
    int temp = 1;
    while (b--) {
      sol[i] = temp;
      i += 2;
      temp++;
    }
    sol[0] = B + 1;
    temp = B + 2;
    int index = A + B + 1;
    if (B == A) {
      while (index != N) {
        sol[index] = temp;
        temp++;
        index++;
      }
    } else {
      temp = N - A;
      while (index != N) {
        sol[index] = temp;
        temp--;
        index++;
      }
    }
    if (flag) {
      for (int i = 0; i < N; i++) {
        sol[i] = N + 1 - sol[i];
      }
    }
    for (int i = 0; i < N; i++) {
      cout << sol[i] << " ";
    }
    cout << "\n";
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, a, b;
    cin >> n >> a >> b;
    int N = n;
    if (abs(a - b) > 1) {
      cout << -1 << "\n";
      continue;
    }
    if (a + b > n - 2) {
      cout << -1 << "\n";
      continue;
    }
    bool flag = false;
    if (a < b) {
      swap(a, b);
      flag = true;
    }
    int A = a;
    int B = b;
    int sol[n];
    int i = 1;
    while (a--) {
      sol[i] = n;
      i += 2;
      n--;
    }
    i = 2;
    int temp = 1;
    while (b--) {
      sol[i] = temp;
      i += 2;
      temp++;
    }
    sol[0] = B + 1;
    temp = B + 2;
    int index = A + B + 1;
    if (B == A) {
      while (index != N) {
        sol[index] = temp;
        temp++;
        index++;
      }
    } else {
      temp = N - A;
      while (index != N) {
        sol[index] = temp;
        temp--;
        index++;
      }
    }
    if (flag) {
      for (int i = 0; i < N; i++) {
        sol[i] = N + 1 - sol[i];
      }
    }
    for (int i = 0; i < N; i++) {
      cout << sol[i] << " ";
    }
    cout << "\n";
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}

```