/* ChampionsNeverFall :)*/
#include <bits/stdc++.h>
using namespace std;
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define repr(i,a,b) for(ll i=a;i>=b;i--)
#define mp make_pair
#define mod 998244353
#define vll vector<ll>
#define vpll vector<pair<ll,ll> >
#define pll pair<ll,ll>
#define f(n) for(ll i=0;i<n;i++)
#define pb push_back
#define iPair pair<ll, pair<ll,ll>>
#define en '\n'
typedef long long ll;
typedef long double ld;
ld PI=2*acos(0.0);
ll pw(ll x,ll y);
int main()
{
    IOS
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    ll tc=1;
    cin>>tc;
    while(tc--)
    {
        ll n;cin>>n;
        //string str;cin>>str;
        vll v(n);f(n)cin>>v[i];
        ll l=0,r=n-1;
        vll ans(n);
        ll i=0;
        while(i<n)
        {
            if(i%2==0)
                ans[i]=v[l++];
            else
                ans[i]=v[r--];
            i++;
        }
        f(n)
        cout<<ans[i]<<" ";
        cout<<en;
    }
    return 0;
}
ll pw(ll x,ll y)
{
    ll temp = 1;
    x = x % mod;
    if (x == 0) return 0;
    while (y > 0)
    {
        if (y & 1)
            temp = (temp*x) % mod;
        y = y>>1;
        x = (x*x) % mod;
    }
    return temp;
}

### Prompt

Create a solution in Cpp for the following 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].

### Response

```cpp
/* ChampionsNeverFall :)*/
#include <bits/stdc++.h>
using namespace std;
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define repr(i,a,b) for(ll i=a;i>=b;i--)
#define mp make_pair
#define mod 998244353
#define vll vector<ll>
#define vpll vector<pair<ll,ll> >
#define pll pair<ll,ll>
#define f(n) for(ll i=0;i<n;i++)
#define pb push_back
#define iPair pair<ll, pair<ll,ll>>
#define en '\n'
typedef long long ll;
typedef long double ld;
ld PI=2*acos(0.0);
ll pw(ll x,ll y);
int main()
{
    IOS
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    ll tc=1;
    cin>>tc;
    while(tc--)
    {
        ll n;cin>>n;
        //string str;cin>>str;
        vll v(n);f(n)cin>>v[i];
        ll l=0,r=n-1;
        vll ans(n);
        ll i=0;
        while(i<n)
        {
            if(i%2==0)
                ans[i]=v[l++];
            else
                ans[i]=v[r--];
            i++;
        }
        f(n)
        cout<<ans[i]<<" ";
        cout<<en;
    }
    return 0;
}
ll pw(ll x,ll y)
{
    ll temp = 1;
    x = x % mod;
    if (x == 0) return 0;
    while (y > 0)
    {
        if (y & 1)
            temp = (temp*x) % mod;
        y = y>>1;
        x = (x*x) % mod;
    }
    return temp;
}

```