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".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline void dump(T& v) {
  for (auto i = v.begin(); i != v.end(); i++) {
    cout << *i << ((i == --v.end()) ? '\n' : ' ');
  }
}
inline string getline() {
  string s;
  getline(cin, s);
  return s;
}
inline void yn(const bool b) {
  b ? cout << ("yes") << '\n' : cout << ("no") << '\n';
}
inline void Yn(const bool b) {
  b ? cout << ("Yes") << '\n' : cout << ("No") << '\n';
}
inline void YN(const bool b) {
  b ? cout << ("YES") << '\n' : cout << ("NO") << '\n';
}
struct io {
  io() {
    ios::sync_with_stdio(false);
    cin.tie(0);
  }
};
const int INF = INT_MAX;
const long long int LLINF = 1LL << 60;
const long long int MOD = 1000000007;
const double EPS = 1e-9;
bool check(vector<int>& v) {
  int max = -1;
  for (int i = 0; i < (v.size() - 1); ++i) {
    if (abs(v[i] - v[i + 1]) < max)
      return 0;
    else {
      max = abs(v[i] - v[i + 1]);
    }
  }
  return 1;
}
int main() {
  int Q;
  cin >> Q;
  while (Q--) {
    int n;
    cin >> n;
    vector<int> v(n), ans1, ans2, v1, v2;
    for (int i = 0; i < (n); ++i) cin >> v[i];
    sort(v.begin(), v.end());
    int pos = ceil(n / 2.0) - 1;
    int diff = 1;
    while (pos < n && pos >= 0) {
      ans1.push_back(v[pos]);
      pos += diff;
      if (diff > 0)
        diff++;
      else
        diff--;
      diff = -diff;
    }
    if (check(ans1))
      dump(ans1);
    else {
      pos = n / 2;
      diff = -1;
      while (pos < n && pos >= 0) {
        ans2.push_back(v[pos]);
        pos += diff;
        if (diff > 0)
          diff++;
        else
          diff--;
        diff = -diff;
      }
      dump(ans2);
    }
  }
}