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:
s = input()
l = list(s)
mid = len(l)//2
if(len(l)%2==0):
    lf = list(s[:mid-1])
    ls = list(s[mid:])
    lf.reverse()
    '''print(lf, ls)
    print(l[mid - 1])'''
    sm = str(l[mid-1])
    fs =  ''
    for i in range(min(len(ls),len(lf))):
        fs = fs + str(ls[i]) +str(lf[i])
    print(sm+fs + str(ls[-1]))
if(len(l)%2!=0):
    lf= list(s[:mid])
    ls = list(s[mid+1:])
    lf.reverse()
    '''print(lf, ls)
    print(l[mid - 1])'''
    sm = str(l[mid])
    fs=''
    for i in range(min(len(ls),len(lf))):
        fs = fs + str(ls[i]) + str(lf[i])
    print(sm+fs)