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 fn(i,n) for(i = 0; i < n; i++)
#define fg(i,a ,n) for(i = a; i < n; i+=2)
#define ft(i,a ,n) for(i = a; i < n; i++)
#define fp(i,a, n) for(i = a; i >= n; i-- )
#define ld double
#define ll long long int
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define pb push_back
#define pf push_front
#define mp make_pair
#define S second
#define F first
#define vi vector <ll>
#define INF 10000000000
#define all(x) x.begin(), x.end()
typedef vector< pair <ll , ll>> vl;
typedef pair<string , string> pairs;
ll i , j, n, k, t,  p = 0,q = 9,a,b, c = 0, ct = 0, m = 0, s, sum = 0, sum1 = 0, sum2 = 0;
int32_t main(){
     cin >> t;
     fn(i , t){
         cin >> n;
         deque <ll> d;
         fn(j , n){
             cin >> a;d.pb(a);
         }
         fn(j , n/2){
             cout << d.front() << " " << d.back() << " ";
             d.pop_front();d.pop_back();
         }
         if(d.size() > 0) cout << d.front() << " ";
         cout << endl;
     }
}