B. Sorted Adjacent Differences

B. Sorted Adjacent Differences

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You have array of nn numbers a1,a2,…,ana1,a2,…,an.

Rearrange these numbers to satisfy |a1−a2|≤|a2−a3|≤…≤|an−1−an||a1−a2|≤|a2−a3|≤…≤|an−1−an|, where |x||x| denotes absolute value of xx. It's always possible to find such rearrangement.

Note that all numbers in aa are not necessarily different. In other words, some numbers of aa may be same.

You have to answer independent tt test cases.

Input

The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The first line of each test case contains single integer nn (3≤n≤1053≤n≤105) — the length of array aa. It is guaranteed that the sum of values of nn over all test cases in the input does not exceed 105105.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (−109≤ai≤109−109≤ai≤109).

Output

For each test case, print the rearranged version of array aa which satisfies given condition. If there are multiple valid rearrangements, print any of them.

Example

input

Copy

2
6
5 -2 4 8 6 5
4
8 1 4 2

output

Copy

5 5 4 6 8 -2
1 2 4 8

Note

In the first test case, after given rearrangement, |a1−a2|=0≤|a2−a3|=1≤|a3−a4|=2≤|a4−a5|=2≤|a5−a6|=10|a1−a2|=0≤|a2−a3|=1≤|a3−a4|=2≤|a4−a5|=2≤|a5−a6|=10. There are other possible answers like "5 4 5 6 -2 8".

In the second test case, after given rearrangement, |a1−a2|=1≤|a2−a3|=2≤|a3−a4|=4|a1−a2|=1≤|a2−a3|=2≤|a3−a4|=4. There are other possible answers like "2 4 8 1".

=========================================================================

排序，先让最小和最大的放在最后，这样一定是满足的

a1 a2  a3....an-1 an

a2 a3  ....an-1         a1  an

再把a2 an-1放在最后， a2 an-1  a1 an

首先a1最小，它和an的差值绝对值一定最大，与an-1的差值绝对值不一定最大，但是一定比a2和an-1的差值绝对值大，然后以此类类推即可

# include
# include
# include
 
using namespace std;
 
int a[100000+10];
int ans[100000+10];
 
int main ()
{
 
    int t;
    cin>>t;
 
    while(t--)
    {
        int n;
        cin>>n;
 
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
 
        sort(a+1,a+1+n);
 
        int left=1,right=n,cnt=0;
 
        int len=n;
        while(left<=right)
        {
            ans[len]=a[right];
            len--;
            if(!len)
            {
                break;
            }
 
            ans[len]=a[left];
            len--;
 
            if(!len)
                break;
 
            left++;
            right--;
 
        }
 
        for(int i=1;i<=n;i++)
        {
            cout<" ";
        }
 
        cout<
 
 
 
    }
    return 0;
 
}