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 fo(i,k,n) for(long long i=k;i<n;i++)
#define endl "\n"
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

typedef long long ll;

/* ll fun_ceil(ll n,ll k){
if(n%k==0){
    return n/k;
}
else {
    ll x = (n/k)+1;
    return x;
}
} */
//sort(s.begin(),s.end(),[](tt x, tt y){return abs(x)>abs(y);});
void calc(){
    int n;cin>>n;
    vector<int> s(n);
    for(int i=0;i<n;i++)
        cin>>s[i];

    int x = 0;
    int y = n-1;
    for(int i=1;i<=n;i++){
        if(i%2==0)
            cout<<s[y--]<<" ";
        else cout<<s[x++]<<" ";
    }
    cout<<endl;
}

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    ll t;cin>>t;

    while(t--){
        calc();
    }
    return 0;
}