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;
const int maxn = 3e5 + 5;
const int mod = 1e9 + 7;
int a[maxn], vis[maxn];
int solve() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  sort(a, a + n);
  int mid = n / 2;
  printf("%d ", a[mid]);
  if (n & 1)
    for (int i = 1; i < mid + 1; ++i) printf("%d %d ", a[mid - i], a[mid + i]);
  else {
    for (int i = 1; i < mid; ++i) printf("%d %d ", a[mid - i], a[mid + i]);
    printf("%d", a[0]);
  }
  return printf("\n");
}
int main() {
  int t;
  for (int i = 0; i < 1e5; ++i) t = i * i, t = sqrt(i), t = pow(2, i);
  int _;
  scanf("%d", &_);
  while (_--) solve();
}