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: Accepted)

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

#define INF 1001001001
#define LINF 1001001001001001001
#define MOD 1000000007
#define MOD2 998244353

#define all(a)  (a).begin(),(a).end()
#define dig(n)  to_string(n).length()

#define pb push_back
#define eb emplace_back
#define mkp make_pair

#define se second
#define fi first



typedef long long ll;
typedef pair<int,int> pii;
typedef pair<pii,pii> ppii;

int main(){

    bool flag=false;
    ll ans=0,sum=0;

    int n;
    cin >>n;
    list<int> l;

    for (int i = 0; i < n; i++){
        int ai;
        cin>>ai;
        
        if(i % 2==0){
            l.push_back(ai);
        }else{
            l.push_front(ai);
        }
    }
    
    if(n%2==1)l.reverse();
    //cout <<fixed<<setprecision(16)<< << endl;
    //cout << ans << endl;

    for(int li : l)cout << li << " ";

    //if(flag)cout << "Yes" <<endl;
    //else cout << "No" <<endl;
    
    return 0;
}