Problem:
Score : 300 points

Problem Statement

You are given an integer sequence of length n, a_1, ..., a_n.
Let us consider performing the following n operations on an empty sequence b.

The i-th operation is as follows:

Append a_i to the end of b.

Reverse the order of the elements in b.

Find the sequence b obtained after these n operations.

Constraints

1 \leq n \leq 2\times 10^5

0 \leq a_i \leq 10^9

n and a_i are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print n integers in a line with spaces in between.
The i-th integer should be b_i.

Sample Input 1

4
1 2 3 4

Sample Output 1

4 2 1 3

After step 1 of the first operation, b becomes: 1.

After step 2 of the first operation, b becomes: 1.

After step 1 of the second operation, b becomes: 1, 2.

After step 2 of the second operation, b becomes: 2, 1.

After step 1 of the third operation, b becomes: 2, 1, 3.

After step 2 of the third operation, b becomes: 3, 1, 2.

After step 1 of the fourth operation, b becomes: 3, 1, 2, 4.

After step 2 of the fourth operation, b becomes: 4, 2, 1, 3.

Thus, the answer is 4 2 1 3.

Sample Input 2

3
1 2 3

Sample Output 2

3 1 2

As shown above in Sample Output 1, b becomes 3, 1, 2 after step 2 of the third operation. Thus, the answer is 3 1 2.

Sample Input 3

1
1000000000

Sample Output 3

1000000000

Sample Input 4

6
0 6 7 6 7 0

Sample Output 4

0 6 6 0 7 7

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <bits/stdc++.h>
     
#define uint unsigned int
#define ll long long
#define ull unsigned long long
#define ld long double
#define rep(i, l, r) for (int i = l; i < r; i++)
#define repb(i, r, l) for (int i = r; i > l; i--)
#define sz(a) (int)a.size()
#define fi first
#define se second
#define mp(a, b) make_pair(a, b)
     
using namespace std;

const int N = 100000;
const int inf = (int)1e9 + 1;
const ll big = (ll)1e18 + 1;
const int P = 31;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 2;

int main()
{
    //freopen("a.in", "r", stdin);
    //freopen("a.out", "w", stdout);
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.precision(20);
    //ll TL = 0.95 * CLOCKS_PER_SEC;
    //clock_t time = clock();
    int n;
    cin >> n;
    deque<int> a;
    rep(i, 0, n)
    {
        int x;
        cin >> x;
        if (i % 2 == 0)
            a.push_back(x);
        else
            a.push_front(x);
    }
    if (n % 2 == 0)
        reverse(a.begin(), a.end());
    for (int i : a)
        cout << i << " ";
    cout << "\n";
    return 0;
}