#include <bits/stdc++.h>
namespace infinities {
const long long maxn = 1e6 + 7;
const long long INF = 1e9 + 7;
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
long long t, n, a, b, p[maxn];
void print() {
  for (register long long i = 1; i <= n; i++) cout << p[i] << " ";
  cout << "\n";
}
signed main() {
  t = read();
  while (t--) {
    n = read(), a = read(), b = read();
    if (abs(a - b) >= 2) {
      cout << "-1\n";
      continue;
    }
    register long long cnt = 0;
    if (a > b) {
      for (register long long i = 1; i <= b + 1; i++) {
        p[++cnt] = n - a * 2 + i, p[++cnt] = n - a * 2 + i + b + 1;
      }
      if (n - cnt < 1) {
        puts("-1");
        continue;
      }
      for (register long long i = cnt + 1; i <= n; i++) p[i] = n - i + 1;
      print();
      continue;
    }
    if (a < b) {
      for (register long long i = 1; i <= a + 1; i++) {
        p[++cnt] = i + b, p[++cnt] = i;
      }
      if (n - cnt < 1) {
        puts("-1");
        continue;
      }
      for (register long long i = cnt + 1; i <= n; i++) p[i] = i;
      print();
      continue;
    }
    if (a == b) {
      for (register long long i = 1; i <= a + 1; i++)
        p[++cnt] = i, p[++cnt] = i + b + 1;
      if (n - cnt < 0) {
        puts("-1");
        continue;
      }
      for (register long long i = cnt + 1; i <= n; i++) p[++cnt] = i;
      print();
      continue;
    }
  }
  return 0;
}
}  // namespace infinities
signed main() { return infinities::main(); }

### 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>
namespace infinities {
const long long maxn = 1e6 + 7;
const long long INF = 1e9 + 7;
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
long long t, n, a, b, p[maxn];
void print() {
  for (register long long i = 1; i <= n; i++) cout << p[i] << " ";
  cout << "\n";
}
signed main() {
  t = read();
  while (t--) {
    n = read(), a = read(), b = read();
    if (abs(a - b) >= 2) {
      cout << "-1\n";
      continue;
    }
    register long long cnt = 0;
    if (a > b) {
      for (register long long i = 1; i <= b + 1; i++) {
        p[++cnt] = n - a * 2 + i, p[++cnt] = n - a * 2 + i + b + 1;
      }
      if (n - cnt < 1) {
        puts("-1");
        continue;
      }
      for (register long long i = cnt + 1; i <= n; i++) p[i] = n - i + 1;
      print();
      continue;
    }
    if (a < b) {
      for (register long long i = 1; i <= a + 1; i++) {
        p[++cnt] = i + b, p[++cnt] = i;
      }
      if (n - cnt < 1) {
        puts("-1");
        continue;
      }
      for (register long long i = cnt + 1; i <= n; i++) p[i] = i;
      print();
      continue;
    }
    if (a == b) {
      for (register long long i = 1; i <= a + 1; i++)
        p[++cnt] = i, p[++cnt] = i + b + 1;
      if (n - cnt < 0) {
        puts("-1");
        continue;
      }
      for (register long long i = cnt + 1; i <= n; i++) p[++cnt] = i;
      print();
      continue;
    }
  }
  return 0;
}
}  // namespace infinities
signed main() { return infinities::main(); }

```