#include <bits/stdc++.h>
using namespace std;
const long long INF = 9e18;
const long long N = 2e5 + 5;
const long long M = 1e9 + 7;
bool chk(vector<long long> a, long long x, long long y) {
  long long A = 0, B = 0;
  for (long long i = 2; i < a.size(); i++)
    if (a[i] > a[i - 1] && a[i] > a[i + 1])
      A++;
    else if (a[i] < a[i - 1] && a[i] < a[i + 1])
      B++;
  if (A == x && B == y) return 1;
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long tests;
  cin >> tests;
  while (tests--) {
    vector<long long> a;
    long long n, x, y;
    bool ok = 0;
    cin >> n >> x >> y;
    if (abs(x - y) > 1 || x + y > n - 2) {
      cout << "-1\n";
      continue;
    }
    if (x >= y) {
      long long l = 1;
      long long r = n;
      while (l <= r) {
        a.push_back(l++);
        if (l > r) break;
        a.push_back(r--);
      }
      long long idx = 1 + 2 * y;
      if (x > y) idx++;
      sort(a.begin() + idx, a.begin() + n);
      if (x > y) reverse(a.begin() + idx, a.begin() + n);
    } else {
      long long l = 1;
      long long r = n;
      while (l <= r) {
        a.push_back(r--);
        if (l > r) break;
        a.push_back(l++);
      }
      long long idx = 2 * y;
      sort(a.begin() + idx, a.begin() + n);
    }
    for (auto h : a) cout << h << " ";
    cout << "\n";
  }
}

### 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;
const long long INF = 9e18;
const long long N = 2e5 + 5;
const long long M = 1e9 + 7;
bool chk(vector<long long> a, long long x, long long y) {
  long long A = 0, B = 0;
  for (long long i = 2; i < a.size(); i++)
    if (a[i] > a[i - 1] && a[i] > a[i + 1])
      A++;
    else if (a[i] < a[i - 1] && a[i] < a[i + 1])
      B++;
  if (A == x && B == y) return 1;
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long tests;
  cin >> tests;
  while (tests--) {
    vector<long long> a;
    long long n, x, y;
    bool ok = 0;
    cin >> n >> x >> y;
    if (abs(x - y) > 1 || x + y > n - 2) {
      cout << "-1\n";
      continue;
    }
    if (x >= y) {
      long long l = 1;
      long long r = n;
      while (l <= r) {
        a.push_back(l++);
        if (l > r) break;
        a.push_back(r--);
      }
      long long idx = 1 + 2 * y;
      if (x > y) idx++;
      sort(a.begin() + idx, a.begin() + n);
      if (x > y) reverse(a.begin() + idx, a.begin() + n);
    } else {
      long long l = 1;
      long long r = n;
      while (l <= r) {
        a.push_back(r--);
        if (l > r) break;
        a.push_back(l++);
      }
      long long idx = 2 * y;
      sort(a.begin() + idx, a.begin() + n);
    }
    for (auto h : a) cout << h << " ";
    cout << "\n";
  }
}

```