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:
a = str(input())
l = list(a)
b = []
if (len(l) == 1) or all(x==l[0] for x in l):
    print(a)
else:    
    c = len(l)//2
    if len(l)%2 == 0:
        b = l[c - 1]
        c = c - 1
    else:
        b = l[c]
    if len(l)%2 == 0:
        for i in range(1, len(l)//2):
            b = b + l[c+i] + l[c-i]
        b = b + l[len(l)-1]
        print(b)
    else:
        for i in range(1, len(l)//2+1):
            b = b + l[c+i] + l[c-i]
        print(b)