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 <iostream>
#include "bits/stdc++.h"
using namespace std;
#define print(x) cout<<x<<endl;
#define rep(i,a,b) for(int i=a;i<b;i++)
#define REP(i,a) for(int i=0;i<a;i++)
#define jkl int debug = 0;
#define RREP(i,a) for(int i=a-1;i>=0;i--)
#define rrep(i,a,b) for(int i=a-1;i>=b;i--)
#define printall(n,array) {for(int i=0;i<n;i++){cout<<array[i]<<" ";}cout<<endl;}
#define U() cout<<endl;
typedef long long ll;
typedef pair<int, int> Pi;
typedef pair<int, Pi> V;
typedef vector<ll> VE;
const ll mod = 1e9 + 7; //10^9+7


int main(){
    ll n;
    ll a[200002];
    cin >> n;
    REP(i, n)cin >> a[i];
    VE ans;
    ll q = n - 1;

    while(q >= 0){
        ans.push_back(a[q]);
        q -= 2;
    }
    if(n % 2){q = 1;}
    else q = 0;
    while(q < n){
        ans.push_back(a[q]);
        q += 2;
    }
    REP(i, ans.size() - 1)cout << ans[i] << " ";
    cout << ans[ans.size() - 1];
}