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>
#include<iostream>
#include <algorithm>
#include <math.h>
#include <vector>
#include<set>
#include<map>
#include<string>
#include<functional>
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define ll long long
#define tc() ll tc;cin>>tc;while(tc--)
#define For(a,b) for(ll i=a;i<b;i++)
#define arrin(arr,n) for(ll i=0;i<n;i++) cin>>arr[i];
#define matin(arr,n,m) for(ll i=0;i<n;i++) for(ll j=0;j<m;j++) cin>>arr[i][j]
#define matout(arr,n,m) for(ll i=0;i<n;i++){ for(ll j=0;j<m;j++) cout<<arr[i][j]<<" ";  cout<<endl;}
#define vect vector<ll>
#define vectout(v) for (int i = 0; i < v.size(); i++) cout<<v[i]<<" "; cout<<endl
#define eb emplace_back
#define pb pop_back
#define vsort(v) sort(v.begin(), v.end())
#define mod 1e9 + 7;
using namespace std;

int main() {
	fast

	
	tc(){
		ll n,c=0;
		cin>>n;
		ll arr[n],brr[n];
		arrin(arr,n);
		For(0,n){
			if(i%2 == 0){
				brr[i]=arr[i/2];
			}
			else{
				c++;
				brr[i] = arr[n-c];
			}
		}
		For(0,n)
		cout<<brr[i]<<" ";
		cout<<endl;
	}

}