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:
// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#include <bits/stdc++.h>
#define ll long long int
#define vi vector<int>
#define vvi vector<vector<int>>
#define vll vector<long long>
#define vs vector<string>
#define vc vector<char>
#define vb vector<bool>
#define forn(i, s, n)  for(ll i=(ll)s; i<(ll)(n); i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define pll pair<long long int, long long int>
#define pii pair<int, int>
#define pss pair<string, string>
#define ull unsigned long long int
#define lld long double
#define F first
#define S second
#define PI 3.141592653589793238
#define prec(n) fixed<<setprecision(n)
#define ordered_set tree<ll, null_type,less_equal<ll>, rb_tree_tag,tree_order_statistics_node_update>
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
#define itsval(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); itval(_it, args); }
using namespace std;
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;

void itval(istream_iterator<string> it) {}
template<typename T, typename... Args>
void itval(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << endl;
    itval(++it, args...);
}
const ll MOD = 1e9 + 7;
template <typename T>
inline void print(T x) {cout << x << "\n";}
template <typename T>
inline void printvec(T x) {for (auto a : x) cout << a << ' '; cout << '\n';}
// -----------------------------------------------------------------------

struct custom {
    bool operator()(const pair<pll, ll> &p1, const pair<pll, ll> &p2)const {
        if (p1.S == p2.S) {
            return p1.F.S < p2.F.S;
        }
        return p1.S < p2.S;
    }
};


// Calculate a^b % MOD -------------------------------------------------
ll get_pow(ll a, ll b) {
    ll res = 1;
    while (b) {
        if (b & 1) res = (res * a) % MOD;
        a = (a * a) % MOD;
        b >>= 1;
    }
    return res;
}
// ---------------------------------------------------------------------
const ll N = 2e5 + 5, inf = 4e18;

void solve()
{
    int n;
    cin >> n;
    vll v(n), ans;
    forn(i, 0, n)
    {
        cin >> v[i];
    }
    int u = 0, d = n - 1;
    forn(i, 0, n / 2)
    {
        ans.pb(v[i]);
        ans.pb(v[n - i - 1]);
    }
    if (n % 2)
        ans.pb(v[n / 2]);
    printvec(ans);
}


int 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 test = 1;
    cin >> test;
    clock_t z = clock();
    forn(tes, 0, test)
    {
        solve();
    }
    debug("Total Time:%.4f\n", (double)(clock() - z) / CLOCKS_PER_SEC);
    return 0;
}