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)

/*
TLE発生
原因はvectorのreverseかな？
*/


#include <bits/stdc++.h>
using namespace std;

using LL = long long;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    LL n;
    cin >> n;
    vector<LL> a(n);
    for (int i = 0; i < n; i++){
        cin >> a[i];
    }

    vector<LL> b;
    b.reserve(n);

    if(n == 1){
        cout << a[0] << endl;
        return 0;
    }

    for (int i = 0; i < n; i = i + 2){
        b.push_back(a[i]);
    }
    reverse(b.begin(), b.end());
    for (int i = 1; i < n; i = i + 2){
        b.push_back(a[i]);
    }
    reverse(b.begin(), b.end());

    for (int i = 0; i < n; i++){
        cout << b[i] << " ";
    }

    cout << endl;

    return 0;
}