
# slower version:

s = input()
for i in range(len(s)):
  if i%2 == 0:
    print(s[i], end = '')


# optimized version of the same code:

s = input()
ans = ''
for i in range(len(s)):
  if i % 2 == 0:
    ans += s[i]
print(ans)



# slower version:

import numpy as np

N = int(eval(input()))

A = np.array(list(map(int, input().split())))

count = 0

while True:

    if np.sum(A%2) > 0:

        break

    count +=1

    A = A//2

print(count)


# optimized version of the same code:

n = int(eval(input()))

a = list(map(int, input().split()))



ans = float("inf")



for i in a:

    ans = min(ans, len(bin(i)) - bin(i).rfind("1") - 1) # (1)…2で割った回数が最小のものを探索

print(ans)
import numpy as np

N = int(eval(input()))
A = np.array(list(map(int, input().split())))
count = 0
while True:
    if np.sum(A % 2) > 0:
        break
    count += 1
    A = A // 2
print(count)

n = int(eval(input()))
a = list(map(int, input().split()))
ans = float("inf")
for i in a:
    ans = min(ans, len(bin(i)) - bin(i).rfind("1") - 1)  # (1)…2で割った回数が最小のものを探索
print(ans)

-import numpy as np - -N = int(eval(input())) -A = np.array(list(map(int, input().split()))) -count = 0 -while True: -    if np.sum(A % 2) > 0: -        break -    count += 1 -    A = A // 2 -print(count) +n = int(eval(input())) +a = list(map(int, input().split())) +ans = float("inf") +for i in a: +    ans = min(ans, len(bin(i)) - bin(i).rfind("1") - 1)  # (1)…2で割った回数が最小のものを探索 +print(ans)