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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>


int main(void){
    int n;
    int *a;

    scanf("%d",&n);

    a=(int *)malloc(sizeof(int)*n);

    for(int i=0;i<n;i++) scanf("%d",&a[i]);

    int *ans;
    ans=(int *)malloc(sizeof(int)*n);

    int m=n-1;
    int k=0;

    if(n%2==0){
        while(m>0){
            ans[k]=a[m];
            k++;
            m-=2;
        }
        m++;
        while(m<n){
            ans[k]=a[m];
            k++;
            m+=2;
        }
    }
    else{
        while(m>=0){
            ans[k]=a[m];
            k++;
            m-=2;
        }
        m+=3;
        while(m<n){

            //printf("::%d,%d\n",a[m],m);

            ans[k]=a[m];
            k++;
            m+=2;
        }
    }

    for(int i=0;i<n;i++) printf("%d ",ans[i]);
    printf("\n");
    return 0;
}