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: Accepted)

#include<stdio.h>

#define N 100000

int main(){
  int odd[N], even[N], i, n;

  scanf("%d", &n);
  if(n==1){
    scanf("%d", &n);
      printf("%d\n", n);
  }else if(n%2==0){
    for(i=0; i<n/2; i++){
      scanf("%d%d", &odd[i], &even[i]);
      //printf("%d\n", odd[1]);
    }
    // printf("%d\n", n/2-1);
    for(i=n/2-1; i>-1; i--){
      printf("%d ", even[i]);
    }
    for(i=0; i<n/2; i++){
      if(i==n/2-1){
	printf("%d\n", odd[i]);
      }else{
	printf("%d ", odd[i]);
      }
    } 
  }else{
    for(i=0; i<n/2; i++){
      scanf("%d%d", &odd[i], &even[i]);
      //printf("%d\n", odd[i]);
    }
    scanf("%d", &odd[i]);
    //printf("%d %d\n", odd[i], i);
    for(i=n/2; i>-1; i--){
      printf("%d ", odd[i]);
    }
    for(i=0; i<n/2; i++){
      if(i==n/2-1){
	printf("%d\n", even[i]);
      }else{
	printf("%d ", even[i]);
      }
    }
  }
  
  return 0;
}