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:
for _ in range(int(input())):
	n = int(input())
	l=list(map(int, input().split()))
	l.sort()
	midi=False
	if(n%2==1):
		midi=True
		mid=l[n//2]
		l=l[:n//2]+l[(n//2)+1:]
		n=n-1
	a=[]
	for i in range(n//2):
		a.append(l[i])
		a.append(l[n-i-1])
	if(midi):
		f=False
		if(abs(a[0]-mid)>=abs(a[1]-a[0])):
			a=[mid]+a
		elif(abs(a[-1]-a[-2])>=abs(a[-1]-mid)):
			a=a+[mid]
		else:	
			for i in range(n-1):
				if(abs(a[i]-mid)>=abs(a[i+1]-mid)):
					a=a[:i+1]+[mid]+a[i+1:]
					f=True
					break
	a=a[::-1]
	print(*a)