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>
using namespace std;
vector<int> inpVec(int n) {
  vector<int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  return v;
}
void printVec(vector<int> v) {
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  int t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v = inpVec(n);
    sort(v.begin(), v.end());
    int start = 0;
    int end = n - 1;
    vector<int> ans;
    while (start < end) {
      ans.push_back(v[end--]);
      ans.push_back(v[start++]);
    }
    if (start == end) ans.push_back(v[start]);
    reverse(ans.begin(), ans.end());
    printVec(ans);
  }
  return 0;
}