#include <bits/stdc++.h>
using namespace std;
int i = 0, j = 0, k;
int main() {
  int t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    int arr[n];
    for (int i = 0; i < n; ++i) cin >> arr[i];
    sort(arr, arr + n);
    if (n & 1) {
      int q = n / 2;
      cout << arr[q] << " ";
      i = q + 1;
      --q;
      bool l = true, r = false;
      while (q >= 0 || i < n) {
        if (l && q > -1) {
          cout << arr[q] << " ";
          l = false;
          r = true;
          --q;
          continue;
        }
        if (r && i < n) {
          cout << arr[i] << " ";
          r = false, l = true;
          ++i;
        }
      }
      cout << endl;
    } else {
      i = (n - 1) / 2;
      j = i + 1;
      bool l = true, r = false;
      while (i > -1 || j < n) {
        if (l && i > -1) {
          cout << arr[i] << " ";
          l = false;
          r = true;
          --i;
          continue;
        }
        if (r && j < n) {
          cout << arr[j] << " ";
          r = false;
          l = true;
          ++j;
        }
      }
      cout << endl;
    }
  }
}

### Prompt

Please formulate a Cpp solution to the following problem:

You have array of n numbers a_{1}, a_{2}, …, a_{n}. 

Rearrange these numbers to satisfy |a_{1} - a_{2}| ≤ |a_{2} - a_{3}| ≤ … ≤ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.

Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.

You have to answer independent t test cases.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^{4}) — the number of test cases.

The first line of each test case contains single integer n (3 ≤ n ≤ 10^{5}) — the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.

The second line of each test case contains n integers a_{1}, a_{2}, …, a_{n} (-10^{9} ≤ a_{i} ≤ 10^{9}).

Output

For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.

Example

Input


2
6
5 -2 4 8 6 5
4
8 1 4 2


Output


5 5 4 6 8 -2
1 2 4 8

Note

In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 ≤ |a_{2} - a_{3}| = 1 ≤ |a_{3} - a_{4}| = 2 ≤ |a_{4} - a_{5}| = 2 ≤ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".

In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 ≤ |a_{2} - a_{3}| = 2 ≤ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int i = 0, j = 0, k;
int main() {
  int t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    int arr[n];
    for (int i = 0; i < n; ++i) cin >> arr[i];
    sort(arr, arr + n);
    if (n & 1) {
      int q = n / 2;
      cout << arr[q] << " ";
      i = q + 1;
      --q;
      bool l = true, r = false;
      while (q >= 0 || i < n) {
        if (l && q > -1) {
          cout << arr[q] << " ";
          l = false;
          r = true;
          --q;
          continue;
        }
        if (r && i < n) {
          cout << arr[i] << " ";
          r = false, l = true;
          ++i;
        }
      }
      cout << endl;
    } else {
      i = (n - 1) / 2;
      j = i + 1;
      bool l = true, r = false;
      while (i > -1 || j < n) {
        if (l && i > -1) {
          cout << arr[i] << " ";
          l = false;
          r = true;
          --i;
          continue;
        }
        if (r && j < n) {
          cout << arr[j] << " ";
          r = false;
          l = true;
          ++j;
        }
      }
      cout << endl;
    }
  }
}

```