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: Wrong Answer)

#include <stdio.h>
int main()
{
    int n,m,i,j,k;
    scanf("%d",&n);
    long a[n],ki[n],gu[n];
    for(i=1;i<n+1;i++)
    {
        scanf("%ld",&a[i]);
    }
    for(m=1;m<n+1;m++)
    {
       if(m%2 == 0)
       {
           gu[m/2]=a[m];
       }
       else if(m%2 ==1)
       {
           ki[(m+1)/2]=a[m];
       }
    }
    if(n%2==0)
    {
        for(j=n/2;j>0;j--)
        {
            printf("%ld ",gu[j]);
        }
        for(j=1;j<n/2+1;j++)
        {
            printf("%ld ",ki[j]);
        }
    }
    else if(n%2==1)
    {
        for(k=1;k<(n+1)/2+1;k++)
        {
            printf("%ld ",ki[k]);
        }
        for(k=(n-1)/2;k>0;k--)
        {
            printf("%ld ",gu[k]);
        }
    }
}