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:
'''
def IsPrime(N):
    int(N)
    for i in range(2, int(N)):
        if int(N)%i == 0:
            N=0
    if int(N)==1:
        N=0
    return N
m, n = map(int, input().split())
k=0
s=[]
for i in range(m, n+1):
    if IsPrime(i)!=0:
        s+=i
        k+=1
for i in range(len(s)):
    print(s[i])
if k == 0:
    print('Absent')

    
'''
'''
def bin_search(lst, n):
    lenth = len(lst)
    x = 0
    while x!=lenth:
        half = (x+lenth)//2
        if lst[half]==n:
            return n+1


a = [int(s) for s in input('Введите массив:').split()]
elem = int(input('Введите искомый элемент'))
print(bin_search(a, elem))
'''
n = int(input())
a = input()
k=1
r=1
while k<=n:
    print(a[k-1], end='')
    k+=r
    r+=1