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<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int a[1000560];
int main(){
	int n,j,i;
	while(~scanf("%d",&n)){
		int t;
		for(i=1;i<=n;i++){
			scanf("%d",&a[i]);
		}
		if(n%2==0){
			for(i=n;i>0;i-=2){
				printf("%d ",a[i]);
			}
			for(i=1;i<=n-1;i+=2){
				if(i==n-1){
					printf("%d",a[i]);
				}
				else{
					printf("%d ",a[i]);
				}	 
			}
		}
		if(n%2==1){
			for(i=n;i>=1;i-=2){
				
				printf("%d ",a[i]);
				
			}for(i=2;i<=n-1;i+=2){
				
				if(i==n-1){
					printf("%d",a[i]);
				}
				else{
					printf("%d ",a[i]);
				}
			}
		}
		printf("\n");
	}
	return 0;
}