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: Runtime Error)

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[100005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        if(n==1) printf("%d\n",a[0]);
        else if(n==2) printf("%d %d\n",a[1],a[0]);
        else
        {
            printf("%d",a[n-1]);
            for(int i=n-3; i>=0; i-=2)
                printf(" %d",a[i]);
            if(n%2==1)
                for(int i=1; i<n-1; i+=2)
                    printf(" %d",a[i]);
            if(n%2==0)
                for(int i=0; i<n-1; i+=2)
                    printf(" %d",a[i]);
            printf("\n");
        }
    }
    return 0;
}