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>
#include <stdlib.h>

int main(void){

	int n,i,i2,j;
	int * a,*b;
	int back = 0;
	int haiti;
	int tmp;

	scanf("%d",&n);
	
	a = (int*)malloc(sizeof(int)*n);
	if(a == NULL){
		printf("失敗しました。\n");
		return 0;
	}
	
	
	b = (int*)malloc(sizeof(int)*n);
	if(b == NULL){
		printf("失敗しました。\n");
		return 0;
	}


	

	for(i = 1; i <= n; i++){
              scanf("%d",&a[i-1]);

		b[back++] = a[i-1];

		haiti = i / 2;
		i2 = i;

		for(j = 0; j < haiti; j++){
			tmp = b[j];
			b[j] = b[--i2];
			b[i2] = tmp;
		}
	}

	for(i = 0; i < n; i++){
		printf("%d ",b[i]);
	}

	return 0;
}