#include <bits/stdc++.h>
using namespace std;
inline void boost() {
  ios::sync_with_stdio(false);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
template <typename T>
inline bool innrange(T lo, T hi, T val) {
  return val >= lo && hi >= val;
}
template <typename T>
inline bool inrange(T lo, T hi, T val) {
  return val > lo && hi > val;
}
static constexpr auto N = ((100ull) * 1024ull);
static constexpr auto MOD = 1000000007;
static constexpr auto INF = numeric_limits<int>::max();
static constexpr auto INFl = numeric_limits<long long>::max();
array<int, N> na;
array<pair<int, int>, N> da;
int main() {
  boost();
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int ni = 0; ni < n; ni++) cin >> na[ni];
    sort(begin(na), begin(na) + n);
    vector<int> res;
    for (int ni = 0; ni <= n / 2; ni++) {
      if (n == (1 + 2 * ni)) {
        res.push_back(na[ni]);
      } else if (ni != n / 2) {
        res.push_back(na[n - ni - 1]);
        res.push_back(na[ni]);
      }
    }
    reverse(begin(res), end(res));
    for (auto i : res) cout << i << " ";
    cout << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
inline void boost() {
  ios::sync_with_stdio(false);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
template <typename T>
inline bool innrange(T lo, T hi, T val) {
  return val >= lo && hi >= val;
}
template <typename T>
inline bool inrange(T lo, T hi, T val) {
  return val > lo && hi > val;
}
static constexpr auto N = ((100ull) * 1024ull);
static constexpr auto MOD = 1000000007;
static constexpr auto INF = numeric_limits<int>::max();
static constexpr auto INFl = numeric_limits<long long>::max();
array<int, N> na;
array<pair<int, int>, N> da;
int main() {
  boost();
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int ni = 0; ni < n; ni++) cin >> na[ni];
    sort(begin(na), begin(na) + n);
    vector<int> res;
    for (int ni = 0; ni <= n / 2; ni++) {
      if (n == (1 + 2 * ni)) {
        res.push_back(na[ni]);
      } else if (ni != n / 2) {
        res.push_back(na[n - ni - 1]);
        res.push_back(na[ni]);
      }
    }
    reverse(begin(res), end(res));
    for (auto i : res) cout << i << " ";
    cout << endl;
  }
  return 0;
}

```