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 PYTHON3):

Solution:
#                                               |
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#  (   |  |   |  (   |  (     (   |  |   |  (   |  (   |
# \__,_| _|  _| \__,_| \___| \___/  _|  _| \__,_| \__,_|

import sys
import collections

def read_line():
	return sys.stdin.readline()[:-1]
 
def read_int():
	return int(sys.stdin.readline())
 
def read_int_line():
	return [int(v) for v in sys.stdin.readline().split()]

t = read_int()
for i in range(t):
	n = read_int()
	a = read_int_line()
	b = list(a)
	b.sort(reverse=True)
	a.sort()

	ans = []

	if n%2==0:
		for i in range(n//2-1,-1,-1):
			ans.append(a[i])
			ans.append(b[i])

	if n%2!=0:
		for i in range(n//2-1,-1,-1):
			ans.append(a[i])
			ans.append(b[i])

		# if abs(ans[-1]-ans[-2]) <= abs(ans[-1]-a[n//2]):
		# 	ans.append(a[n//2])
		# else:
		ans = [a[n//2]] + ans
	
	print(*ans)