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:
t = input()
l = list(t)
l1 = []
l2 = []
if len(t) == 1 or len(t) == 2:
    print(t)
else:
    if len(t) % 2 == 0:
        x = len(t)//2 - 1
    else:
        x = len(t)//2
    for i in range(x+1, len(t)):
        l1.append(t[i])
    for i in range(0, x):
        l2.append(t[i])
    print(t[x],end="")
    k = x-1
    m = x+1
    while k >= 0 and m < len(t):
        print("{}{}".format(t[m], t[k]), end="")
        m+=1
        k-=1
    if len(t) % 2 == 0:
        print(t[len(t)-1],end="")