#include <bits/stdc++.h>
using namespace std;
int sgn(int x) {
  if (x < 0) return -1;
  return 1;
}
int main() {
  int T;
  cin >> T;
  for (int t = 1; t <= T; t++) {
    int n;
    cin >> n;
    int a, b;
    cin >> a >> b;
    if (abs(a - b) > 1 || a + b > n - 2) {
      cout << -1 << endl;
      continue;
    }
    vector<int> A;
    A.push_back(0);
    if (a >= b) {
      for (int i = 1; i <= a; i++) {
        A.push_back(i);
        A.push_back(-i);
      }
      if (a == b) {
        A.push_back(a + 1);
      }
    } else if (a < b) {
      for (int i = 1; i <= b; i++) {
        A.push_back(-i);
        A.push_back(i);
      }
    }
    int dir = 1;
    if (A.size() >= 2) {
      dir = sgn(A[A.size() - 1] - A[A.size() - 2]);
    }
    while (A.size() < n) {
      A.push_back(A.back() + dir);
    }
    set<int> uniq;
    for (int a : A) {
      uniq.insert(a);
    }
    map<int, int> ids;
    for (int x : uniq) {
      int sz = ids.size();
      ids[x] = sz + 1;
    }
    if (A.size() != n) {
      cout << -1 << endl;
      continue;
    }
    for (int& a : A) {
      a = ids[a];
    }
    for (int i = 1; i <= n - 2; i++) {
      a -= (A[i] > max(A[i - 1], A[i + 1]));
      b -= (A[i] < min(A[i - 1], A[i + 1]));
    }
    if (a != 0 || b != 0) {
      cout << -1 << endl;
      continue;
    }
    for (int& a : A) {
      printf("%d ", a);
    }
    cout << endl;
  }
}

### Prompt

Develop a solution in cpp 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;
int sgn(int x) {
  if (x < 0) return -1;
  return 1;
}
int main() {
  int T;
  cin >> T;
  for (int t = 1; t <= T; t++) {
    int n;
    cin >> n;
    int a, b;
    cin >> a >> b;
    if (abs(a - b) > 1 || a + b > n - 2) {
      cout << -1 << endl;
      continue;
    }
    vector<int> A;
    A.push_back(0);
    if (a >= b) {
      for (int i = 1; i <= a; i++) {
        A.push_back(i);
        A.push_back(-i);
      }
      if (a == b) {
        A.push_back(a + 1);
      }
    } else if (a < b) {
      for (int i = 1; i <= b; i++) {
        A.push_back(-i);
        A.push_back(i);
      }
    }
    int dir = 1;
    if (A.size() >= 2) {
      dir = sgn(A[A.size() - 1] - A[A.size() - 2]);
    }
    while (A.size() < n) {
      A.push_back(A.back() + dir);
    }
    set<int> uniq;
    for (int a : A) {
      uniq.insert(a);
    }
    map<int, int> ids;
    for (int x : uniq) {
      int sz = ids.size();
      ids[x] = sz + 1;
    }
    if (A.size() != n) {
      cout << -1 << endl;
      continue;
    }
    for (int& a : A) {
      a = ids[a];
    }
    for (int i = 1; i <= n - 2; i++) {
      a -= (A[i] > max(A[i - 1], A[i + 1]));
      b -= (A[i] < min(A[i - 1], A[i + 1]));
    }
    if (a != 0 || b != 0) {
      cout << -1 << endl;
      continue;
    }
    for (int& a : A) {
      printf("%d ", a);
    }
    cout << endl;
  }
}

```