#include <bits/stdc++.h>
const double PI = 3.1415926535897932384626433;
const int KL = 1e6;
const long long MOD = 1e9 + 7;
using namespace std;
int q, x, y, z, n, m, a, b, ans;
string t;
vector<int> vec;
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &n);
    scanf("%d", &a);
    scanf("%d", &b);
    if (a + b <= n - 2 && abs(a - b) <= 1) {
      if (a == b + 1) {
        int cnt = n - 1;
        while (a > 0) {
          printf("%d %d ", cnt, cnt + 1);
          cnt -= 2;
          a -= 1;
        }
        for (int i = cnt + 1; i >= 1; i--) printf("%d ", i);
        printf("\n");
      } else if (a == b) {
        int cnt = a + b;
        while (a > 0) {
          printf("%d %d ", cnt, cnt + 1);
          cnt -= 2;
          a--;
        }
        printf("1 ");
        for (int i = b + b + 2; i <= n; i++) printf("%d ", i);
        printf("\n");
      } else {
        int cnt = 2;
        while (b > 0) {
          printf("%d %d ", cnt, cnt - 1);
          cnt += 2;
          b--;
        }
        for (int i = cnt - 1; i <= n; i++) printf("%d ", i);
        printf("\n");
      }
    } else
      printf("-1\n");
  }
  return 0;
}

### Prompt

Generate 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>
const double PI = 3.1415926535897932384626433;
const int KL = 1e6;
const long long MOD = 1e9 + 7;
using namespace std;
int q, x, y, z, n, m, a, b, ans;
string t;
vector<int> vec;
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &n);
    scanf("%d", &a);
    scanf("%d", &b);
    if (a + b <= n - 2 && abs(a - b) <= 1) {
      if (a == b + 1) {
        int cnt = n - 1;
        while (a > 0) {
          printf("%d %d ", cnt, cnt + 1);
          cnt -= 2;
          a -= 1;
        }
        for (int i = cnt + 1; i >= 1; i--) printf("%d ", i);
        printf("\n");
      } else if (a == b) {
        int cnt = a + b;
        while (a > 0) {
          printf("%d %d ", cnt, cnt + 1);
          cnt -= 2;
          a--;
        }
        printf("1 ");
        for (int i = b + b + 2; i <= n; i++) printf("%d ", i);
        printf("\n");
      } else {
        int cnt = 2;
        while (b > 0) {
          printf("%d %d ", cnt, cnt - 1);
          cnt += 2;
          b--;
        }
        for (int i = cnt - 1; i <= n; i++) printf("%d ", i);
        printf("\n");
      }
    } else
      printf("-1\n");
  }
  return 0;
}

```