#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, a, b;
    cin >> n >> a >> b;
    if (abs(a - b) > 1 || a + b > n - 2) {
      cout << -1 << endl;
    } else {
      int flip = 0;
      if (a < b) swap(a, b), flip = 1;
      vector<int> ans(n);
      int l = 0, r = 1;
      for (int i = 0; i < n - a - b; i++) ans[i] = r++;
      for (int i = 0; i < a + b; i++) {
        if (i % 2 == 0) {
          ans[n - a - b + i] = l--;
        } else {
          ans[n - a - b + i] = r++;
        }
      }
      for (int i = 0; i < n; i++) ans[i] -= l;
      if (flip) {
        for (int& v : ans) v = n + 1 - v;
        swap(a, b);
      }
      for (int v : ans) cout << v << ' ';
      cout << endl;
      set<int> seen;
      int mi = 0, ma = 0;
      for (int i = 0; i < n; i++) {
        if (ans[i] < 1 || ans[i] > n || seen.count(ans[i])) assert(0);
        if (i > 0 && i + 1 < n) {
          ma += ans[i] > ans[i - 1] && ans[i] > ans[i + 1];
          mi += ans[i] < ans[i - 1] && ans[i] < ans[i + 1];
        }
      }
      assert(ma == a);
      assert(mi == b);
    }
  }
}

### Prompt

Please create a solution in cpp 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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, a, b;
    cin >> n >> a >> b;
    if (abs(a - b) > 1 || a + b > n - 2) {
      cout << -1 << endl;
    } else {
      int flip = 0;
      if (a < b) swap(a, b), flip = 1;
      vector<int> ans(n);
      int l = 0, r = 1;
      for (int i = 0; i < n - a - b; i++) ans[i] = r++;
      for (int i = 0; i < a + b; i++) {
        if (i % 2 == 0) {
          ans[n - a - b + i] = l--;
        } else {
          ans[n - a - b + i] = r++;
        }
      }
      for (int i = 0; i < n; i++) ans[i] -= l;
      if (flip) {
        for (int& v : ans) v = n + 1 - v;
        swap(a, b);
      }
      for (int v : ans) cout << v << ' ';
      cout << endl;
      set<int> seen;
      int mi = 0, ma = 0;
      for (int i = 0; i < n; i++) {
        if (ans[i] < 1 || ans[i] > n || seen.count(ans[i])) assert(0);
        if (i > 0 && i + 1 < n) {
          ma += ans[i] > ans[i - 1] && ans[i] > ans[i + 1];
          mi += ans[i] < ans[i - 1] && ans[i] < ans[i + 1];
        }
      }
      assert(ma == a);
      assert(mi == b);
    }
  }
}

```