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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)


#include<stdio.h>
int MAX=2e5;
 
int main()
{
	int n;
	while(scanf("%d",&n))
	{
		long a[MAX+1],b[MAX+1];
		
		for(int i=0;i<n;i++)
			scanf("%ld",&a[i]);
		
		if(n&1)
        {
            for(int i=n-1;i>=0;i--)
            {
            	if(!(i&1))
            	{
            		printf("%d ",a[i]);
				}
			}
			for(int i=0;i<=n-1;i++)
            {
            	if(i&1)
            	{
            		printf("%d ",a[i]);
				}	
			}
			printf("\n");
        }
		else
        {
            for(int i=n-1;i>=1;i--)
            {
            	if(i&1)
            	{
            		printf("%d ",a[i]);
				}
			}
			for(int i=0;i<=n-1;i++)
            {
            	if(!(i&1))
            	{
            		printf("%d ",a[i]);
				}
			}
			printf("\n");
        }	
	}
	
	
	return 0;
}

/*
#include<iostream>
using namespace std;
const int MAXN=2e5+5;
int a[MAXN];


int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=1;i<=n;++i)   
        {
            cin>>a[i];
        }
        if(n&1)
        {
            cout<<a[n];
            for(int i=n-2;i>=1;i-=2)
                cout<<" "<<a[i];
            for(int i=2;i<=n-1;i+=2)
                cout<<" "<<a[i];
            cout<<endl;
        }else
        {
            cout<<a[n];
            for(int i=n-2;i>=2;i-=2)
                cout<<" "<<a[i];
            for(int i=1;i<=n-1;i+=2)
                cout<<" "<<a[i];
            cout<<endl;
        }
    }
    return 0;
}
*/