#include <bits/stdc++.h>
using namespace std;
class Solution {
 public:
  static bool check(const vector<int> &a) {
    for (int i = 1; i + 1 < a.size(); ++i) {
      if (std::abs(a[i] - a[i - 1]) > std::abs(a[i + 1] - a[i])) {
        return false;
      }
    }
    return true;
  }
  void arrange(vector<int> &a) const {
    sort(a.begin(), a.end());
    vector<int> ret;
    ret.reserve(a.size());
    int left = 0, right = a.size() - 1;
    while (left <= right) {
      ret.push_back(a[left++]);
      if (left <= right) {
        ret.push_back(a[right--]);
      }
    }
    reverse(ret.begin(), ret.end());
    a = ret;
  }
  void solve(int n) {
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
    }
    arrange(a);
    for (int i = 0; i < n; ++i) {
      printf("%d ", a[i]);
    }
    cout << endl;
  }
  void gen() {
    for (int iter = 0; iter < 40; ++iter) {
      int n = rand() % 30;
      vector<int> a(n);
      for (int i = 0; i < n; ++i) {
        a[i] = rand() % 50;
      }
      arrange(a);
      cout << endl;
      if (!check(a)) {
        for (int i = 0; i < n; ++i) {
          cout << a[i] << " ";
        }
        exit(1);
      }
    }
  }
};
int main(int, char **) {
  int tests;
  cin >> tests;
  while (tests--) {
    int n;
    cin >> n;
    Solution solution;
    solution.solve(n);
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
class Solution {
 public:
  static bool check(const vector<int> &a) {
    for (int i = 1; i + 1 < a.size(); ++i) {
      if (std::abs(a[i] - a[i - 1]) > std::abs(a[i + 1] - a[i])) {
        return false;
      }
    }
    return true;
  }
  void arrange(vector<int> &a) const {
    sort(a.begin(), a.end());
    vector<int> ret;
    ret.reserve(a.size());
    int left = 0, right = a.size() - 1;
    while (left <= right) {
      ret.push_back(a[left++]);
      if (left <= right) {
        ret.push_back(a[right--]);
      }
    }
    reverse(ret.begin(), ret.end());
    a = ret;
  }
  void solve(int n) {
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
    }
    arrange(a);
    for (int i = 0; i < n; ++i) {
      printf("%d ", a[i]);
    }
    cout << endl;
  }
  void gen() {
    for (int iter = 0; iter < 40; ++iter) {
      int n = rand() % 30;
      vector<int> a(n);
      for (int i = 0; i < n; ++i) {
        a[i] = rand() % 50;
      }
      arrange(a);
      cout << endl;
      if (!check(a)) {
        for (int i = 0; i < n; ++i) {
          cout << a[i] << " ";
        }
        exit(1);
      }
    }
  }
};
int main(int, char **) {
  int tests;
  cin >> tests;
  while (tests--) {
    int n;
    cin >> n;
    Solution solution;
    solution.solve(n);
  }
  return 0;
}

```