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;
pair<long long int, long long int> getfc(double x) {
  long long int y = (long long int)x;
  long long int z = (long long int)x;
  if (x - (double)y > 0) {
    z++;
  }
  return make_pair(y, z);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int T, n, m, k, i, j, l;
  vector<long long int> a, b;
  string anss;
  cin >> T;
  while (T--) {
    cin >> n;
    a.resize(n);
    b.resize(0);
    b.resize(n);
    for (i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a.begin(), a.end());
    j = n - 1;
    k = n - 1;
    l = 0;
    for (i = 0; i < n / 2; i++) {
      b[j--] = a[k--];
      b[j--] = a[l++];
    }
    if (n % 2 != 0) {
      b[j--] = a[k];
    }
    for (i = 0; i < n; i++) {
      cout << b[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}