1339_B. Sorted Adjacent Differences_10803
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".
for _ in range(int(input())):
	n = int(input())
	ls = sorted(list(map(int, input().split())))
	mid = n // 2
	if n % 2 == 0:
		mid -= 1
	i = mid - 1
	turn = True
	j = mid + 1
	ar = []
	ar.append(ls[mid])
	while i >= 0 or j < n:
		if turn and j < n:
			ar.append(ls[j])
			j += 1
		elif not turn and i >= 0:
			ar.append(ls[i])
			i -= 1
		turn = not turn
	print(*ar)

Below is a problem description that describes the problem. Write code in Python that appropriately solves the problem.

### Description:
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".

### Input:
2
6
5 -2 4 8 6 5
4
8 1 4 2

### Output:
5 5 4 6 -2 8
2 4 1 8


### Input:
1
29
31 16383 15 127 255 3 4095 536870911 7 63 262143 2097151 1 16777215 134217727 524287 511 8388607 67108863 1023 8191 2047 32767 33554431 268435455 65535 131071 1048575 4194303

### Output:
32767 65535 16383 131071 8191 262143 4095 524287 2047 1048575 1023 2097151 511 4194303 255 8388607 127 16777215 63 33554431 31 67108863 15 134217727 7 268435455 3 536870911 1


### Code:
for _ in range(int(input())):
	n = int(input())
	ls = sorted(list(map(int, input().split())))
	mid = n // 2
	if n % 2 == 0:
		mid -= 1
	i = mid - 1
	turn = True
	j = mid + 1
	ar = []
	ar.append(ls[mid])
	while i >= 0 or j < n:
		if turn and j < n:
			ar.append(ls[j])
			j += 1
		elif not turn and i >= 0:
			ar.append(ls[i])
			i -= 1
		turn = not turn
	print(*ar)
