#include <iostream>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <algorithm>
#include <iomanip>
#include <set>
using namespace std;

/*ϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕ*/

#define int long long int
#define ld long double
#define F first
#define S second
#define P pair <int,int>
#define vi vector <int>
#define vs vector <string>
#define vb vector <bool>
#define all(x) x.begin(),x.end()
#define sz(x) (int)x.size()
#define REP(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define REV(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define sp(x,y) fixed<<setprecision(y)<<x
#define pb push_back
#define endl '\n'
const int mod = 1e9 + 7;

/*ϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕ*/

const int N = 1e5 + 5;

void solve() {

	int n; cin >> n;
	int a[n + 1];
	REP(i, 1, n)
	cin >> a[i];

	int i = 1;
	int it = n;
	int x = 0;
	vi ans;
	while (i < n) {
		ans.pb(a[i]);
		x++;
		if (x > n)
			break;
		ans.pb(a[n - i + 1]);
		//cout << a[n - i + 1] << " ";
		i++;
		it -= 2;
		if (it == 0)
			break;

	}
	if (sz(ans) >= 1)
		for (int i = 0; i < n; i++) {
			cout << ans[i] << " ";
		}
	else
		cout << a[1];

	cout << endl;

	return ;
}

int32_t main() {

	/* → → → → → → → → → → → → → → → → → → → → → → → → → → → →
	→ → → → → → → → → → → → → → → → → → → → → → → → → → → → */

	ios_base:: sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);

#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	/* → → → → → → → → → → → → → → → → → → → → → → → → → → → →
	→ → → → → → → → → → → → → → → → → → → → → → → → → → → → */

	int t; cin >> t; while (t--)
		solve();

	return 0;
}
### Prompt

Your task is to create a cpp solution to 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
#include <iostream>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <algorithm>
#include <iomanip>
#include <set>
using namespace std;

/*ϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕ*/

#define int long long int
#define ld long double
#define F first
#define S second
#define P pair <int,int>
#define vi vector <int>
#define vs vector <string>
#define vb vector <bool>
#define all(x) x.begin(),x.end()
#define sz(x) (int)x.size()
#define REP(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define REV(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define sp(x,y) fixed<<setprecision(y)<<x
#define pb push_back
#define endl '\n'
const int mod = 1e9 + 7;

/*ϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕϕ*/

const int N = 1e5 + 5;

void solve() {

	int n; cin >> n;
	int a[n + 1];
	REP(i, 1, n)
	cin >> a[i];

	int i = 1;
	int it = n;
	int x = 0;
	vi ans;
	while (i < n) {
		ans.pb(a[i]);
		x++;
		if (x > n)
			break;
		ans.pb(a[n - i + 1]);
		//cout << a[n - i + 1] << " ";
		i++;
		it -= 2;
		if (it == 0)
			break;

	}
	if (sz(ans) >= 1)
		for (int i = 0; i < n; i++) {
			cout << ans[i] << " ";
		}
	else
		cout << a[1];

	cout << endl;

	return ;
}

int32_t main() {

	/* → → → → → → → → → → → → → → → → → → → → → → → → → → → →
	→ → → → → → → → → → → → → → → → → → → → → → → → → → → → */

	ios_base:: sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);

#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	/* → → → → → → → → → → → → → → → → → → → → → → → → → → → →
	→ → → → → → → → → → → → → → → → → → → → → → → → → → → → */

	int t; cin >> t; while (t--)
		solve();

	return 0;
}
```