
# 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 sys



def main():

    input = sys.stdin.readline

    N = int(eval(input()))

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



    ans = 0

    for C in range(1, N//2):

        n = N // C

        point = 0

        used = set([0])

        for k in range(1, n):

            A = N - 1 - k*C



            if A in used or k*C in used or A == k*C:

                break

            used.add(A)

            used.add(k*C)



            point += S[A] + S[k*C]

            ans = max(ans, point)



    return ans





if __name__ == '__main__':

    print((main()))


# optimized version of the same code:

import sys



def main():

    input = sys.stdin.readline

    N = int(eval(input()))

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



    ans = 0

    # C = A - B

    for C in range(1, N//2):

        n = (N-1) // C

        point = 0

        for k in range(1, n):

            A = N-1 - k*C



            # Check A and B are positive integer, and S_i is not used twice.

            #

            # B = A - C and B >= 0, so A must be over C (A > C).

            #

            # If C is aliquot of N-1 (N-1 % C = 0), A is also aliquot of N-1

            # (N-1 % A = 0) because A + kC = N-1. Therefore when A is less than

            # or equal kC, the A's value is already used by kC.

            if A < C or ((N-1) % C == 0 and A <= k*C):

                break



            point += S[A] + S[k*C]

            ans = max(ans, point)



    return ans





if __name__ == '__main__':

    print((main()))

import sys


def main():
    input = sys.stdin.readline
    N = int(eval(input()))
    S = list(map(int, input().split()))
    ans = 0
    for C in range(1, N // 2):
        n = N // C
        point = 0
        used = set([0])
        for k in range(1, n):
            A = N - 1 - k * C
            if A in used or k * C in used or A == k * C:
                break
            used.add(A)
            used.add(k * C)
            point += S[A] + S[k * C]
            ans = max(ans, point)
    return ans


if __name__ == "__main__":
    print((main()))

import sys


def main():
    input = sys.stdin.readline
    N = int(eval(input()))
    S = list(map(int, input().split()))
    ans = 0
    # C = A - B
    for C in range(1, N // 2):
        n = (N - 1) // C
        point = 0
        for k in range(1, n):
            A = N - 1 - k * C
            # Check A and B are positive integer, and S_i is not used twice.
            #
            # B = A - C and B >= 0, so A must be over C (A > C).
            #
            # If C is aliquot of N-1 (N-1 % C = 0), A is also aliquot of N-1
            # (N-1 % A = 0) because A + kC = N-1. Therefore when A is less than
            # or equal kC, the A's value is already used by kC.
            if A < C or ((N - 1) % C == 0 and A <= k * C):
                break
            point += S[A] + S[k * C]
            ans = max(ans, point)
    return ans


if __name__ == "__main__":
    print((main()))

+    # C = A - B -        n = N // C +        n = (N - 1) // C -        used = set([0]) -            if A in used or k * C in used or A == k * C: +            # Check A and B are positive integer, and S_i is not used twice. +            # +            # B = A - C and B >= 0, so A must be over C (A > C). +            # +            # If C is aliquot of N-1 (N-1 % C = 0), A is also aliquot of N-1 +            # (N-1 % A = 0) because A + kC = N-1. Therefore when A is less than +            # or equal kC, the A's value is already used by kC. +            if A < C or ((N - 1) % C == 0 and A <= k * C): -            used.add(A) -            used.add(k * C)