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: Time Limit Exceeded)

#include <bits/stdc++.h>  

using namespace std;

#define FOR(i, j, k) for(int i = j; i < k; ++i)
#define rep(i, j) FOR(i, 0, j)
#define FORr(i, j, k) for(int i = j; i >= k; --i)
#define repr(i, j) FOR(i, j, 0)
#define INF (1 << 30)

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<P, int> Pi;

const ll MOD = 1e9 + 7;
const int dy[]={0, 0, 1, -1};
const int dx[]={1, -1, 0, 0};

template <class T> void chmin(T& a, const T& b) { a = min(a, b); }
template <class T> void chmax(T& a, const T& b) { a = max(a, b); }

vector<int> ans;

int main() {
	int n;
	scanf("%d", &n);
	rep(i, n) {
		int a;
		scanf("%d", &a);
		ans.push_back(a);
		reverse(ans.begin(), ans.end());
	}
	rep(i, n) (i != n - 1) ? printf("%d ", ans[i]) : printf("%d\n", ans[i]);
	return 0;
}