human
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.
"""Te ... Write me a Python program to find a string s such that the above algorithm results in string t?
assistant
#Here is Python program to find a string s such that the above algorithm results in string t.
mplate for Python Competitive Programmers prepared by Mayank Chaudhary """

# to use the print and division function of Python3
from __future__ import division, print_function

"""value of mod"""
MOD = 998244353
mod = 10**9 + 7

"""use resource"""
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])

"""for factorial"""

# def prepare_factorial():
#     fact = [1]
#     for i in range(1, 2005):
#         fact.append((fact[-1] * i) % mod)
#     ifact = [0] * 2005
#     ifact[2004] = pow(fact[2004], mod - 2, mod)
#     for i in range(2004, 0, -1):
#         ifact[i - 1] = (i * ifact[i]) % mod
#
#     return fact, ifact

"""uncomment next 4 lines while doing recursion based question"""
# import threading
# threading.stack_size(1<<27)
import sys
# sys.setrecursionlimit(10000)

"""uncomment modules according to your need"""
# from bisect import bisect_left, bisect_right, insort
# import itertools
# from math import floor, ceil, sqrt, degrees, atan, pi, log
# from heapq import heappop, heapify, heappush
# from random import randint as rn
# from Queue import Queue as Q
# from collections import Counter, defaultdict, deque
# from copy import deepcopy
'''
def modinv(n, p):
    return pow(n, p - 2, p)
'''

#
# def ncr(n, r,  fact, ifact):  # for using this uncomment the lines calculating fact and ifact
#     t = (fact[n] * (ifact[r]*ifact[n-r]) % mod)% mod
#     return t


def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()

"""*****************************************************************************************"""

# def GCD(x, y):
#     while (y):
#         x, y = y, x % y
#     return x

# def lcm(x, y):
#     return (x*y)//(GCD(x, y))

# def get_xor(n):
#     return [n,1,n+1,0][n%4]

# def pow(a, b, d):
#     # print(a, b, d)
#     res = 1
#     while b:
#         if b&1:
#             res = (res%d * a%d)%d
#            ...