#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, a, b;
  cin >> n >> a >> b;
  if (abs(a - b) > 1) {
    cout << -1 << '\n';
    return;
  }
  if (a + b > n - 2) {
    cout << -1 << '\n';
    return;
  }
  long long c = 0;
  long long d = n - 1;
  vector<long long> res;
  if (a >= b) {
    res = {0};
    c++;
    while (a || b) {
      if (a > 0) {
        res.push_back(d);
        a--;
        d--;
      }
      if (b > 0) {
        res.push_back(c);
        b--;
        c++;
      }
    }
  } else {
    res = {n - 1};
    d--;
    while (a || b) {
      if (b > 0) {
        res.push_back(c);
        b--;
        c++;
      }
      if (a > 0) {
        res.push_back(d);
        a--;
        d--;
      }
    }
  }
  if (res.back() == c - 1) {
    for (long long j = c; j <= d; j++) res.push_back(j);
  } else {
    for (long long j = d; j >= c; j--) res.push_back(j);
  }
  for (long long x : res) cout << (x + 1) << ' ';
  cout << '\n';
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) solve();
  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;
void solve() {
  long long n, a, b;
  cin >> n >> a >> b;
  if (abs(a - b) > 1) {
    cout << -1 << '\n';
    return;
  }
  if (a + b > n - 2) {
    cout << -1 << '\n';
    return;
  }
  long long c = 0;
  long long d = n - 1;
  vector<long long> res;
  if (a >= b) {
    res = {0};
    c++;
    while (a || b) {
      if (a > 0) {
        res.push_back(d);
        a--;
        d--;
      }
      if (b > 0) {
        res.push_back(c);
        b--;
        c++;
      }
    }
  } else {
    res = {n - 1};
    d--;
    while (a || b) {
      if (b > 0) {
        res.push_back(c);
        b--;
        c++;
      }
      if (a > 0) {
        res.push_back(d);
        a--;
        d--;
      }
    }
  }
  if (res.back() == c - 1) {
    for (long long j = c; j <= d; j++) res.push_back(j);
  } else {
    for (long long j = d; j >= c; j--) res.push_back(j);
  }
  for (long long x : res) cout << (x + 1) << ' ';
  cout << '\n';
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```