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 <bits/stdc++.h>
#include <cmath>
#include <numeric>
using namespace std;
#define rep(i,a,b) for(int64_t i=(a); i<(b); ++i) // a ≦ i < b 
#define Rrep(i,a,b) for(int64_t i=(a);i>=(b);--i) // reverse repeat. a から b まで減少．
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend() //逆イテレータ
#define RANGE(a,b,c) (a).begin()+b,(a).begin()+c // [b,c) イテレータ
#define MOD 1000000007
#define INF 1000000000
typedef pair<int64_t, int64_t> PII;
typedef vector<int64_t> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef vector<PII> VP;

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

int n; cin >>n ;
VI a(n); rep(i, 0, n) cin >> a[i];

for (int i = n - 1; i > 0; i -= 2)
    cout << a[i] << " ";
for (int i = 0; i < n; i += 2)
    cout << a[i] << " ";

cout << endl;
}
// 書いて考える
// 境界，出力文字列 チェック
// 可読性優先．高速化次点．
// まずは全探索，分割統治，次にDP
// 制限を見る．境界に注意．求めたい量の変域．
// 偶奇，逆から，ソート，出現回数，出現位置，DP， 余事象，包除
// データ構造． 問題の特徴量．単調性，二分探索
// 存在判定：構成方法，入力の特徴
// gcd, lcm ,素因数分解．
// 例外を十分に含む一般化．想像力の限界
// 小さい系から例示
// 代数的処理
// 始めは過剰に例示・場合分けしてもいい．各場合を確実に対処．
// 自明な例から処理，除外．
// 小数のときは，精度の設定する．doubel 変数に数値を入力するときは 123. とする．
// テストケース作成は数表あり