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++ (GCC 9.2.1) (judge verdict: Accepted)

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

typedef long lint;
typedef long long llint;
typedef pair<int, int> pint;
typedef pair<long long, long long> pllint;

// static const int MAX = 1e6;
// static const int NIL = -1;
// static const llint INF = 1<<21;
// static const llint MOD = 1e9 + 7;

bool compPair(const pint& arg1, const pint& arg2) { return arg1.first > arg2.first; }
template<class T> void chmax(T& a, T b) { if (a < b) { a = b; } }
template<class T> void chmin(T& a, T b) { if (a > b) { a = b; } }

int main(void) {
    int n;
    cin >> n;

    list<int> l;
    bool flgpush = true;
    bool flgreverse = true;

    int a;
    for(int in=0;in<n;in++) {
        cin >> a;
        
        if(flgpush) {
            l.push_front(a);
        }
        else {
            l.push_back(a);
        }

        flgpush = !flgpush;
        flgreverse = !flgreverse;
    }

    if(flgreverse) l.reverse();
    for(auto itr=l.begin();itr!=l.end();itr++) {
        cout << *itr << " ";
    }
    cout << endl;
    return 0;
}