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>
#include <stdlib.h>

int main(){
	int i, n, k;
	int *a, *b;

	a = (int *)malloc(200001 * sizeof(int));
	b = (int *)malloc(200001 * sizeof(int));

	scanf("%d", &n);

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

	
	if(n == 1) {
		printf("%d", a[1]);
	}

	else if(n % 2 == 0){
		i = 1;
		while(n - 2 * (i - 1) != 2) {
			b[i] = a[n - 2 * (i - 1)];
			i++;
		}

		b[i] = a[2];
		i++;
		k = 0;

		while(2 * k + 1 != n - 1) {
			b[i] = a[2 * k + 1];
			i++;
			k++;
		}
		b[i] = a[n - 1];

	}

	else if(n % 2 == 1){
		i = 1;
		while(n - 2 * (i - 1) != 1) {
			b[i] = a[n - 2 * (i - 1)];
			i++;
		}

		b[i] = a[1];
		i++;
		k = 1;

		while(2 * k != n - 1) {
			b[i] = a[2 * k];
			i++;
			k++;
		}
		b[i] = a[n - 1];

	}

	for(i = 1; i <= n; i++) {
		printf("%d ", b[i]);
	}
	free(a);
	free(b);   
}