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;
long long int ara[1000005], ara2[1000005];
map<long long int, long long int> vis1;
map<long long int, long long int> vis2;
priority_queue<long long int> pq;
priority_queue<long long int> pq2;
vector<long long int> vec;
queue<long long int> vec1;
int main() {
  long long int n, item;
  scanf("%lld", &n);
  for (long long int i = 1; i <= n; i++) {
    scanf("%lld", &item);
    for (long long int j = 1; j <= item; j++) {
      scanf("%lld", &ara[j]);
    }
    sort(ara + 1, ara + item + 1);
    long long int m = item / 2 + item % 2;
    long long int n = m + 1;
    for (long long int k = 1; k <= item / 2; k++) {
      cout << ara[m] << " " << ara[n] << " ";
      n += 1;
      m -= 1;
    }
    if (item % 2 != 0) cout << ara[1];
    cout << endl;
  }
  return 0;
}