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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

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

int main() {
	int n;
	scanf("%d", &n);
	
	long int *a, *b, *c;

	a = (long int *)malloc(n*sizeof(long int));
	b = (long int *)malloc(n*sizeof(long int));
	c = (long int *)malloc(n*sizeof(long int));

	for (int i = 0; i < n; i++) {
		int count = 0;

		scanf("%d", &a[i]);

		b[i] = a[i];

		for (int j = i; j >= 0; j--) {
			c[count] = b[j];
			count++;
		}
		for (int j = 0; j <= i; j++) {
			b[j] = c[j];
		}
	}

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