A string s of length n can be encrypted by the following algorithm:

  * iterate over all divisors of n in decreasing order (i.e. from n to 1), 
  * for each divisor d, reverse the substring s[1 ... d] (i.e. the substring which starts at position 1 and ends at position d). 



For example, the above algorithm applied to the string s="codeforces" leads to the following changes: "codeforces" → "secrofedoc" → "orcesfedoc" → "rocesfedoc" → "rocesfedoc" (obviously, the last reverse operation doesn't change the string because d=1).

You are given the encrypted string t. Your task is to decrypt this string, i.e., to find a string s such that the above algorithm results in string t. It can be proven that this string s always exists and is unique.

Input

The first line of input consists of a single integer n (1 ≤ n ≤ 100) — the length of the string t. The second line of input consists of the string t. The length of t is n, and it consists only of lowercase Latin letters.

Output

Print a string s such that the above algorithm results in t.

Examples

Input

10
rocesfedoc


Output

codeforces


Input

16
plmaetwoxesisiht


Output

thisisexampletwo


Input

1
z


Output

z

Note

The first example is described in the problem statement.
"""Problem B - Reversing Encryption.

http://codeforces.com/contest/999/problem/b

A string `s` of length `n` can be encrypted by the following algorithm:

- iterate over all divisors of `n` in decreasing order (i.e. from `n` to `1`),

- for each divisor `d`, reverse the substring `s[1 ... d]` (i.e. the substring
which starts at position `1` and ends at position `d`).

For example, the above algorithm applied to the string `s`="codeforces" leads
to the following changes: "codeforces" `\to` "secrofedoc" `\to` "orcesfedoc"
`\to` "rocesfedoc" `\to` "rocesfedoc" (obviously, the last reverse operation
doesn't change the string because `d=1`).

You are given the encrypted string `t`. Your task is to decrypt this string,
i.e., to find a string `s` such that the above algorithm results in string `t`.
It can be proven that this string `s` always exists and is unique.

Input:

The first line of input consists of a single integer `n` (`1 <= n <= 100`) —
the length of the string `t`. The second line of input consists of the string
`t`. The length of `t` is `n`, and it consists only of lowercase Latin letters.

Output:

Print a string `s` such that the above algorithm results in `t`.

"""
# import logging


# fmt = '%(levelname)s - %(name)s (line:%(lineno)s) - %(message)s'
# formatter = logging.Formatter(fmt)

# ch = logging.StreamHandler()
# ch.setLevel(logging.ERROR)
# ch.setFormatter(formatter)

# logger = logging.getLogger('reversing_encryption')
# logger.setLevel(logging.ERROR)
# logger.addHandler(ch)


def solve(n, s):
    chars = list(s)
    for d in [x for x in range(1, n + 1) if n / x == n // x]:
        chars[:d] = reversed(chars[:d])
    return ''.join(chars)


def main():
    n = int(input().strip())
    s = input().strip()

    result = solve(n, s)
    print(result)


if __name__ == '__main__':
    main()
