#include <bits/stdc++.h>
using namespace std;
int N, A, B;
int main(void) {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d", &N, &A, &B);
    if (abs(A - B) > 1) {
      printf("-1\n");
      continue;
    }
    if (A * 2 + 1 > N || B * 2 + 1 > N || A + B > N - 2) {
      printf("-1\n");
      continue;
    }
    if (A >= B) {
      for (int i = 0; i < A; i++) {
        printf("%d %d ", i + 1, N - i);
      }
      if (A == B) {
        for (int i = A + 1; i < N - A + 1; i++) {
          printf("%d%c", i, " \n"[i == N - A]);
        }
      } else {
        for (int i = N - A; i > A; i--) {
          printf("%d%c", i, " \n"[i == A + 1]);
        }
      }
    } else {
      for (int i = 0; i < B; i++) {
        printf("%d %d ", N - i, i + 1);
      }
      for (int i = B + 1; i < N - B + 1; i++) {
        printf("%d%c", i, " \n"[i == N - B]);
      }
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 N, A, B;
int main(void) {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d", &N, &A, &B);
    if (abs(A - B) > 1) {
      printf("-1\n");
      continue;
    }
    if (A * 2 + 1 > N || B * 2 + 1 > N || A + B > N - 2) {
      printf("-1\n");
      continue;
    }
    if (A >= B) {
      for (int i = 0; i < A; i++) {
        printf("%d %d ", i + 1, N - i);
      }
      if (A == B) {
        for (int i = A + 1; i < N - A + 1; i++) {
          printf("%d%c", i, " \n"[i == N - A]);
        }
      } else {
        for (int i = N - A; i > A; i--) {
          printf("%d%c", i, " \n"[i == A + 1]);
        }
      }
    } else {
      for (int i = 0; i < B; i++) {
        printf("%d %d ", N - i, i + 1);
      }
      for (int i = B + 1; i < N - B + 1; i++) {
        printf("%d%c", i, " \n"[i == N - B]);
      }
    }
  }
  return 0;
}

```