#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const long long maxn = 2e5 + 1000;
long long t, n, m, a, b, si[maxn];
signed main() {
  t = read();
  while (t--) {
    n = read();
    a = read();
    b = read();
    long long dis = abs(a - b);
    if (a + b + 2 > n || dis > 1)
      cout << -1 << "\n";
    else {
      if (a > b) {
        for (long long i = 1; i <= a; ++i) cout << i << ' ' << n - i + 1 << ' ';
        long long l = a + 1, r = n + 1 - l;
        for (long long i = r; i >= l; --i) cout << i << ' ';
        cout << "\n";
      } else if (a == b) {
        for (long long i = 1; i <= a; ++i) cout << i << ' ' << n - i + 1 << ' ';
        long long l = a + 1, r = n + 1 - l;
        for (long long i = l; i <= r; ++i) cout << i << ' ';
        cout << "\n";
      } else {
        for (long long i = 1; i <= b; ++i) cout << n - i + 1 << ' ' << i << ' ';
        long long l = b + 1, r = n + 1 - l;
        for (long long i = l; i <= r; ++i) cout << i << ' ';
        cout << "\n";
      }
    }
  }
}

### Prompt

Please formulate 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>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const long long maxn = 2e5 + 1000;
long long t, n, m, a, b, si[maxn];
signed main() {
  t = read();
  while (t--) {
    n = read();
    a = read();
    b = read();
    long long dis = abs(a - b);
    if (a + b + 2 > n || dis > 1)
      cout << -1 << "\n";
    else {
      if (a > b) {
        for (long long i = 1; i <= a; ++i) cout << i << ' ' << n - i + 1 << ' ';
        long long l = a + 1, r = n + 1 - l;
        for (long long i = r; i >= l; --i) cout << i << ' ';
        cout << "\n";
      } else if (a == b) {
        for (long long i = 1; i <= a; ++i) cout << i << ' ' << n - i + 1 << ' ';
        long long l = a + 1, r = n + 1 - l;
        for (long long i = l; i <= r; ++i) cout << i << ' ';
        cout << "\n";
      } else {
        for (long long i = 1; i <= b; ++i) cout << n - i + 1 << ' ' << i << ' ';
        long long l = b + 1, r = n + 1 - l;
        for (long long i = l; i <= r; ++i) cout << i << ' ';
        cout << "\n";
      }
    }
  }
}

```