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:
str=input()

mm=len(str)-1
ss=0
st=[len(str)]

for i in range(len(str)):
    st.append(0)

if len(str)>1: 
    if len(str)%2==0:
                for i in range(0,int(len(str)),2):  
                    if len(str)-1==i:
                        break  
                    st[i]=str[mm]
                    st[i+1]=str[ss]
                    mm-=1
                    ss+=1
                
            
    else:
        for i in range(0,int(len(str)),2):  
                    st[i]=str[ss]
                    if len(str)-1==i:
                        break
                    st[i+1]=str[mm]
                    mm-=1
                    ss+=1
                
            

    for i in reversed(st): 
                    if i!=0:
                        print(i,end='')
else:
    print(str)