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>
using namespace std;
typedef long double LD; 
typedef long long ll; 
#define int ll 
#define ff(i,a,b) for (int i = a; i < b; i++) 
#define bf(i,a,b) for (int i = a; i >= b; i--) 
#define all(v) v.begin(),v.end() 
#define show(a) for(auto xyz:a)cout<<xyz<<" ";cout<<endl;
#define F first 
#define S second
#define pb push_back
#define mp make_pair
#define lb lower_bound   
#define ub upper_bound
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define one cout<<"-1"<<endl
#define pi 3.141592653589793238
#define intmx INT_MAX 
#define intmi INT_MIN
const int MOD=1e9+7;
int mod(int x){
    return ((x%MOD + MOD)%MOD);
}
int add(int a,int b){
    return mod(mod(a)+mod(b));
}
int mul(int a,int b){
    return mod(mod(a)*mod(b));
}

void solve(){
   int n;
   cin>>n;
   int a[n];
   ff(i,0,n)
   cin>>a[i];
   int l=0,r=n-1,f=0;
   ff(i,0,n){
       if(f==0){
           cout<<a[l]<<" ";
           l++;
           f=1;
       }
       else{
           cout<<a[r]<<" ";
           r--;
           f=0;
       }
   }
   cout<<endl;
}
signed main() {
    int t=1;
    cin>>t;
    while(t--){
        solve();
    }
}