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:
import sys
import io, os
import math
def main():
    #input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
    t=int(input())
    for i in range(t):
        n=int(input())
        a=[int(i) for i in input().split()]
        a.sort()
        ans=[]
        if n%2==0:
            for i in range(n//2):
                ans.append(a[n//2-i-1])
                ans.append(a[n//2+i])
        else:
            if (a[n//2]-a[n//2-1])>(a[n//2+1]-a[n//2]):
                for i in range(n//2):
                    ans.append(a[n//2-i])
                    ans.append(a[n//2+i+1])
                ans.append(a[0])
            else:
                for i in range(n//2):
                    ans.append(a[n//2+i])
                    ans.append(a[n//2-i-1])
                    
                ans.append(a[-1])
        print(*ans)

            
 
if __name__ == "__main__": 
    main()