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;
void Fast_IO() {
  ios::sync_with_stdio(0);
  cin.tie(0);
}
int main() {
  Fast_IO();
  int n_testcases;
  cin >> n_testcases;
  for (int n_testcase = 1; n_testcase <= n_testcases; n_testcase++) {
    long long n;
    cin >> n;
    long long a[n];
    for (long long i = (long long)0; i <= (long long)n - 1; i++) cin >> a[i];
    sort(a, a + n);
    vector<long long> ans;
    if (n % 2) {
      ans.push_back(a[n / 2]);
      for (long long i = (long long)0; i <= (long long)n / 2 - 1; i++)
        ans.push_back(a[n / 2 - 1 - i]), ans.push_back(a[n / 2 + 1 + i]);
    } else {
      ans.push_back(a[n / 2 - 1]);
      ans.push_back(a[n / 2]);
      for (long long i = (long long)0; i <= (long long)n / 2 - 2; i++)
        ans.push_back(a[n / 2 - 2 - i]), ans.push_back(a[n / 2 + 1 + i]);
    }
    for (auto x : ans) cout << x << " ";
    cout << '\n';
  }
  return 0;
}