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;
#define endl "\n"
#define int long long 
#define rep(i,j,k) for(int i=j;i<k;i+=1)
#define rrep(i,j,k) for(int i=n-1;i>=0;i-=1)
#define sorti(v) sort(v.begin(),v.end())
#define sortd(v) sort(v.begin(),v.end(),greater<int>())
#define mm map<int,int>
#define vv vector<int>
#define tmm map<int,map<int,int>>
#define trav(a, x) for(auto& a : x)
#define ff first
#define ss second
#define lb lower_bound
#define ub upper_bound
#define all(v) v.begin(),v.end()
#define pb push_back
#define alot(x,y) memset(x,y,sizeof(x))
#define pii pair<int,int>
#define code ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
const int MOD=1e9+7;
const double pi=acos(-1.0);
//memset(array_name,value_to_be_intialized,sizeof(array_name));
// __builtin_popcount(n) count no of set bit in number n
//next_permutation(s.begin(),s.end());
/*int change(string g){
     int p=10;
     int ans=g[g.length()-1]-'0';
     for(int i=g.length()-2;i>=0;i--){
         int j=g[i]-'0';
         ans=p*ans+j;
     }
     return ans;
}
int binpow(int a,int b){
    int res=1;
    while(b>0){
        if(b&1)res*=a;
        a*=a;
        b>>=1;
        
    }
    return res;
}
// lower_bound==> if the element is present it return its first occurence else 1st element which is greater than 
// upper_bound==> it will always return iterator to element which is greater than searched element
*/
int32_t main ()
{
     code;
    int tt;
    cin>>tt;
    //tt=1;
    while(tt--){
        int n;
        cin>>n;
        int a[n+1];
        rep(i,1,n+1)cin>>a[i];
            mm k;
            int p=1;
            int val=ceil(double(n)/2.);
            for(int i=1;i<=val;i++){
                k[p]=a[i];
                p+=2;
            }
            p=2;
            for(int i=n;i>val;i--){
                k[p]=a[i];
                p+=2;
            }
            for(int i=1;i<=n;i++){
                cout<<k[i]<<" ";
            }
            cout<<"\n";

    }
    return 0;
    
}