#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5 + 5;
const int M = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3fll;
const double pi = acos(-1.0);
int read() {
  int a = 0;
  int d = 1;
  char c;
  while (c = getchar(), c < '0' || c > '9')
    if (c == '-') d = -1;
  a = a * 10 + c - 48;
  while (c = getchar(), c >= '0' && c <= '9') a = a * 10 + c - 48;
  a *= d;
  return a;
}
int aa[N];
signed main() {
  int t = read();
  while (t--) {
    int n = read();
    int a = read();
    int b = read();
    swap(a, b);
    for (int i = 1; i <= n; i++) aa[i] = i;
    if (abs(a - b) > 1 || a + b > n - 2) {
      puts("-1");
      continue;
    }
    int id;
    if (a < b) {
      swap(aa[n - 1], aa[n]);
      b--;
      id = 2;
    } else if (a > b) {
      swap(aa[1], aa[2]);
      a--;
      id = 3;
    } else
      id = 2;
    for (int i = 1; i <= a; i++) {
      swap(aa[id], aa[id + 1]);
      id += 2;
    }
    for (int i = 1; i <= n; i++) printf("%d ", aa[i]);
    puts("");
  }
  return 0;
}

### 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5 + 5;
const int M = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3fll;
const double pi = acos(-1.0);
int read() {
  int a = 0;
  int d = 1;
  char c;
  while (c = getchar(), c < '0' || c > '9')
    if (c == '-') d = -1;
  a = a * 10 + c - 48;
  while (c = getchar(), c >= '0' && c <= '9') a = a * 10 + c - 48;
  a *= d;
  return a;
}
int aa[N];
signed main() {
  int t = read();
  while (t--) {
    int n = read();
    int a = read();
    int b = read();
    swap(a, b);
    for (int i = 1; i <= n; i++) aa[i] = i;
    if (abs(a - b) > 1 || a + b > n - 2) {
      puts("-1");
      continue;
    }
    int id;
    if (a < b) {
      swap(aa[n - 1], aa[n]);
      b--;
      id = 2;
    } else if (a > b) {
      swap(aa[1], aa[2]);
      a--;
      id = 3;
    } else
      id = 2;
    for (int i = 1; i <= a; i++) {
      swap(aa[id], aa[id + 1]);
      id += 2;
    }
    for (int i = 1; i <= n; i++) printf("%d ", aa[i]);
    puts("");
  }
  return 0;
}

```