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 Python (3.4.3) (judge verdict: Accepted)

n = int(input())
a = list(map(int,input().split()))

o=[]
e=[]
for i in range(n):
    if i%2==0:
        o.append(a[i])
    if i%2==1:
        e.append(a[i])

if n%2==0:
    E=[]
    for i in range(n//2):
        E.append(e[n//2-1-i])
    ans = []
    for i in range(n//2):
        ans.append(E[i])

    for i in range(n//2):
        ans.append(o[i])
    s = ""
    for i in range(len(ans)):
        s = s+str(ans[i])
        s = s+" "
    print(s)
    exit()
else:
    O=[]
    for i in range(len(o)):
        O.append(o[len(o)-1-i])
    ans = []
    for i in range(len(O)):
        ans.append(O[i])
    for i in range(n//2):
        ans.append(e[i])
    s = ""
    for i in range(len(ans)):
        s = s+str(ans[i])
        s = s+" "
    print(s)
    exit()