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<cstdio>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<algorithm>
#include<math.h>
#include<numeric>
#include<iomanip>

using namespace std;

typedef vector<int> vi;
typedef pair<int, int> pii;
typedef long long ll;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;

const ll INF = 1e15;
const ll MOD = 1e9 + 7;

int main(){
	ll n;
	cin >> n;
	vll A;
	ll i, j;
	for (i = 0; i < n; i++){
		ll a;
		cin >> a;
		A.push_back(a);
	}
	vll B(n, 0);
	if (n % 2 == 0){
		ll num = n - 1;
		for (i = 0; i < n / 2; i++){
			B[i] = A[num];
			num--;
			B[n - i - 1] = A[num];
			num--;
		}
	}
	else{
		ll num = n - 1;
		for (i = 0; i < (n - 1) / 2; i++){
			B[i] = A[num];
			num--;
			B[n - i - 1] = A[num];
			num--;
		}
		B[(n - 1) / 2] = A[num];
	}
	for (i = 0; i < n; i++){
		cout << B[i] << " ";
	}
	cout << endl;
}