Problem: Polycarp has a favorite sequence a[1 ... n] consisting of n integers. He wrote it out on the whiteboard as follows:

  * he wrote the number a_1 to the left side (at the beginning of the whiteboard); 
  * he wrote the number a_2 to the right side (at the end of the whiteboard); 
  * then as far to the left as possible (but to the right from a_1), he wrote the number a_3; 
  * then as far to the right as possible (but to the left from a_2), he wrote the number a_4; 
  * Polycarp continued to act as well, until he wrote out the entire sequence on the whiteboard. 

<image> The beginning of the result looks like this (of course, if n ≥ 4).

For example, if n=7 and a=[3, 1, 4, 1, 5, 9, 2], then Polycarp will write a sequence on the whiteboard [3, 4, 5, 2, 9, 1, 1].

You saw the sequence written on the whiteboard and now you want to restore Polycarp's favorite sequence.

Input

The first line contains a single positive integer t (1 ≤ t ≤ 300) — the number of test cases in the test. Then t test cases follow.

The first line of each test case contains an integer n (1 ≤ n ≤ 300) — the length of the sequence written on the whiteboard.

The next line contains n integers b_1, b_2,…, b_n (1 ≤ b_i ≤ 10^9) — the sequence written on the whiteboard.

Output

Output t answers to the test cases. Each answer — is a sequence a that Polycarp wrote out on the whiteboard.

Example

Input


6
7
3 4 5 2 9 1 1
4
9 2 7 1
11
8 4 3 1 2 7 8 7 9 4 2
1
42
2
11 7
8
1 1 1 1 1 1 1 1


Output


3 1 4 1 5 9 2 
9 1 2 7 
8 2 4 4 3 9 1 7 2 8 7 
42 
11 7 
1 1 1 1 1 1 1 1 

Note

In the first test case, the sequence a matches the sequence from the statement. The whiteboard states after each step look like this:

[3] ⇒ [3, 1] ⇒ [3, 4, 1] ⇒ [3, 4, 1, 1] ⇒ [3, 4, 5, 1, 1] ⇒ [3, 4, 5, 9, 1, 1] ⇒ [3, 4, 5, 2, 9, 1, 1].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define int long long
#define en "\n"
#define fio ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL)
using namespace std;
typedef pair<int, int> pii;

vector<int> sieve(int m){
    vector<bool> is_prime(m+1, 1);
    is_prime[0]=0, is_prime[1]=0;
    vector<int> primes;
    vector<int> factors(m+1);
    for(int i=2; i<=m; i++){
        if(is_prime[i]){
          primes.push_back(i);
          factors[i] = i;
        }
        for(int j=0; j<primes.size() && i*primes[j]<=m && primes[j]<=factors[i]; j++){
          is_prime[i*primes[j]]=0;
          factors[i*primes[j]] = primes[j];
        }
    }
    return primes;
}



int32_t main()
{
	fio;
	int t;
	cin>>t;
	while(t--){
	    int n;
        cin>>n;
        int ar[n];
        for(int i=0; i<n; i++){
            cin>>ar[i];
        }
        int i=0, j=n-1;
        bool flag=0;
        vector<int> res;
        while(i<=j){
            if(flag==0){
                res.push_back(ar[i]);
                i++;
                flag=1;
            }
            else{
                res.push_back(ar[j]);
                j--;
                flag=0;
            }
        }
        for(auto r:res) cout<<r<<" ";
        cout<<en;
	}
		
		
	return 0;
}