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:
'''
# Петя и строки
s = input().lower()
s1 = input().lower()
if s > s1:
    print(1)
elif s < s1:
    print(-1)
else:
    print(0)
# Футбол
s = input()
if '1111111' in s or '0000000' in s:
    print('YES')
else:
    print('NO')
# Юра и заселение
n = int(input())
k = 0
for i in range(n):
    p, q = map(int, input().split())
    if q-p >= 2:
        k += 1
print(k)

# Перевод
s = input()
t = input()
if s[::-1] != t:
    print('NO')
else:
    print('YES')
# Халк
n = int(input())
for i in range(n):
    if i % 2 == 0:
        print('I hate', end = ' ')
    else:
        print('I love', end = ' ')
    if i == n-1:
        print('it', end = ' ')
    else:
        print('that', end = ' ')
# Быстрый математик
n = input()
m = input()
ans = ''
for i in range(len(n)):
    if n[i] == m[i]:
        ans += '0'
    else:
        ans += '1'
print(ans)

# Фотографии Брейна
n, a = map(int, input().split())
color = 'CMY'
bw = 'WGB'
for i in range(n):
    s = input()
    if 'C' in s or 'M' in s or 'Y' in s:
        print('#Color')
        break
else:
    print('#Black&White')
    
# Клавиатура
rl = input()
vvod = input()
ans = ''
a = 'qwertyuiopasdfghjkl;zxcvbnm,./'
if rl == 'R':
    for i in range(len(vvod)):
            ans += a[a.find(vvod[i])-1]
elif rl == 'L':
    for i in range(len(vvod)):
        ans += a[a.find(vvod[i])+1]

print(ans)
'''
# Повторное шифрование
n = int(input())
t = input()
s = ''
c = 1
i = 0
while i < n:
    s += t[i]
    i += c
    c += 1
print(s)