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<cstring>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define maxn 200500
#define inf (int)1e9
#define ll long long
#define mm 1000000007
#define eps 1e-9
#define low(x) x&(-x)
#define pb push_back
#define mp make_pair
#define PII pair<int,int>
using namespace std;

ll read(){
    ll x=0,f=1; char ch=getchar();
    while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while (isdigit(ch)){x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}

int main(){
//	freopen("in.txt","r",stdin);
//	freopen("out.txt","w",stdout);
	int n=read();
	int op=0;
	deque<int> q;
	rep(i,1,n) {
		int x=read();
		if (!op) q.push_front(x);
		else q.push_back(x);
		op^=1; 
	}
	if (!op) {
		while (!q.empty()){
			int x=q.back(); q.pop_back();
			printf("%d ",x);
		}
	}
	else {
		while (!q.empty()) {
			int x=q.front(); q.pop_front();
			printf("%d ",x);
		}
	}
    return 0;
}