#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
const int inf = 1e9;
const int N = 3e5 + 5;
bool flag;
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& i : v) is >> i;
  return is;
}
template <typename T>
void mul(T& a, T b) {
  a += b;
  if (b < (T)0) a += MOD;
  if (a >= (T)MOD) a -= MOD;
}
void solve() {
  int n, a, b;
  cin >> n >> a >> b;
  if (abs(a - b) > 1 or a + b + 2 > n) {
    cout << -1;
    return;
  }
  if (a == 0 and b == 0) {
    for (int i = 1; i <= n; i++) cout << i << ' ';
    return;
  }
  vector<int> ans;
  int l = 1, r = n;
  if (a > b) {
    ans.push_back(l++);
    for (int i = 1; i <= a + b; i++) {
      if (i % 2)
        ans.push_back(r--);
      else
        ans.push_back(l++);
    }
    for (int i = a + b + 1; i < n; i++) ans.push_back(r--);
  } else {
    ans.push_back(r--);
    for (int i = 1; i <= a + b; i++) {
      if (i % 2 == 0)
        ans.push_back(r--);
      else
        ans.push_back(l++);
    }
    for (int i = a + b + 1; i < n; i++) ans.push_back(a == b ? r-- : l++);
  }
  for (auto now : ans) cout << now << ' ';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  int tt = 1;
  cin >> tt;
  while (tt--) {
    flag = 0;
    solve();
    cout << '\n';
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 long long INF = 1e18;
const long long MOD = 1e9 + 7;
const int inf = 1e9;
const int N = 3e5 + 5;
bool flag;
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& i : v) is >> i;
  return is;
}
template <typename T>
void mul(T& a, T b) {
  a += b;
  if (b < (T)0) a += MOD;
  if (a >= (T)MOD) a -= MOD;
}
void solve() {
  int n, a, b;
  cin >> n >> a >> b;
  if (abs(a - b) > 1 or a + b + 2 > n) {
    cout << -1;
    return;
  }
  if (a == 0 and b == 0) {
    for (int i = 1; i <= n; i++) cout << i << ' ';
    return;
  }
  vector<int> ans;
  int l = 1, r = n;
  if (a > b) {
    ans.push_back(l++);
    for (int i = 1; i <= a + b; i++) {
      if (i % 2)
        ans.push_back(r--);
      else
        ans.push_back(l++);
    }
    for (int i = a + b + 1; i < n; i++) ans.push_back(r--);
  } else {
    ans.push_back(r--);
    for (int i = 1; i <= a + b; i++) {
      if (i % 2 == 0)
        ans.push_back(r--);
      else
        ans.push_back(l++);
    }
    for (int i = a + b + 1; i < n; i++) ans.push_back(a == b ? r-- : l++);
  }
  for (auto now : ans) cout << now << ' ';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  int tt = 1;
  cin >> tt;
  while (tt--) {
    flag = 0;
    solve();
    cout << '\n';
  }
  return 0;
}

```