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;
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
void solve() {
  long long n;
  cin >> n;
  long long ret[n + 1], result[n + 1];
  for (long long i = 0; i < n; i++) {
    cin >> ret[i];
  }
  sort(ret, ret + n);
  long long pos = n - 1;
  for (long long i = 0; i < n / 2; i++) {
    result[pos] = ret[n - i - 1];
    pos--;
    result[pos] = ret[i];
    pos--;
  }
  if (n % 2 != 0) {
    result[pos] = ret[n / 2];
  }
  for (long long i = 0; i < n; i++) {
    cout << result[i] << " ";
  }
  cout << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  cin >> t;
  for (long long i = 1; i < t + 1; i++) {
    solve();
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}