Problem: Polycarp loves ciphers. He has invented his own cipher called repeating.

Repeating cipher is used for strings. To encrypt the string s=s_{1}s_{2} ... s_{m} (1 ≤ m ≤ 10), Polycarp uses the following algorithm:

  * he writes down s_1 ones, 
  * he writes down s_2 twice, 
  * he writes down s_3 three times, 
  * ... 
  * he writes down s_m m times. 



For example, if s="bab" the process is: "b" → "baa" → "baabbb". So the encrypted s="bab" is "baabbb".

Given string t — the result of encryption of some string s. Your task is to decrypt it, i. e. find the string s.

Input

The first line contains integer n (1 ≤ n ≤ 55) — the length of the encrypted string. The second line of the input contains t — the result of encryption of some string s. It contains only lowercase Latin letters. The length of t is exactly n.

It is guaranteed that the answer to the test exists.

Output

Print such string s that after encryption it equals t.

Examples

Input


6
baabbb


Output


bab

Input


10
ooopppssss


Output


oops

Input


1
z


Output


z

Solution (in PYTHON3):

Solution:
cifer = input()
res = ''

n = len(cifer)//2

def decifer(s):
    global res
    l = len(s)
    if l == 0:
        return
    if l == 1:
        res += s
        return
    if l % 2 != 0:
        res += s[l//2] + s[l//2 + 1]
        decifer(s[:l//2] + s[l//2 + 2:])
    else:
        res += s[l//2-1] + s[l//2]
        decifer(s[:l//2 - 1] + s[l//2 + 1:])

decifer(cifer)
print(res)


# if len(cifer) % 2 != 0:
#     s = cifer[n]
#     for i in range(n):
#         s += cifer[n+i-1] + cifer[n-i-1]
#     print(s)
# else:
#     s = cifer[n-1] + cifer[n]
#     for i in range(n-1):
#         s = cifer[n+i] + cifer[n-i-1]
#     print(s)