#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1);
int T;
int n, a, b;
bool check() {
  if (a + b > n - 2) return false;
  if (abs(a - b) > 1) return false;
  int t = ceil(1.0 * (n - 2) / 2);
  if (a > t || b > t) return false;
  return true;
}
void print() {
  if (a > b) {
    int i = 1, j = n;
    int t = a;
    while (t--) printf("%d %d ", i++, j--);
    while (i <= j) printf("%d ", j--);
    puts("");
  } else if (b > a) {
    int i = 1, j = n;
    int t = b;
    while (t--) printf("%d %d ", j--, i++);
    while (i <= j) printf("%d ", i++);
    puts("");
  } else {
    int i = 1, j = n;
    int t = a;
    while (t--) printf("%d %d ", i++, j--);
    while (i <= j) printf("%d ", i++);
    puts("");
  }
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d", &n, &a, &b);
    if (!check()) {
      puts("-1");
      continue;
    }
    print();
  }
  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;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1);
int T;
int n, a, b;
bool check() {
  if (a + b > n - 2) return false;
  if (abs(a - b) > 1) return false;
  int t = ceil(1.0 * (n - 2) / 2);
  if (a > t || b > t) return false;
  return true;
}
void print() {
  if (a > b) {
    int i = 1, j = n;
    int t = a;
    while (t--) printf("%d %d ", i++, j--);
    while (i <= j) printf("%d ", j--);
    puts("");
  } else if (b > a) {
    int i = 1, j = n;
    int t = b;
    while (t--) printf("%d %d ", j--, i++);
    while (i <= j) printf("%d ", i++);
    puts("");
  } else {
    int i = 1, j = n;
    int t = a;
    while (t--) printf("%d %d ", i++, j--);
    while (i <= j) printf("%d ", i++);
    puts("");
  }
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d%d", &n, &a, &b);
    if (!check()) {
      puts("-1");
      continue;
    }
    print();
  }
  return 0;
}

```