human
Can you provide a Python code to find a string s such that the above algorithm results in string t given the following:
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.
Using python3 can you solve the prior task?
assistant
#Here is Python code to find a string s such that the above algorithm results in string t.

#        Codeforces Round #490 (Div. 3)
import collections
from functools import cmp_to_key
#key=cmp_to_key(lambda x,y: 1 if x not in y else -1 )
import math
import sys
def getIntList():
    return list(map(int, input().split()))    

import bisect 
try :
    import numpy
    dprint = print
    dprint('debug mode')
except ModuleNotFoundError:
    def dprint(*args, **kwargs):
        pass
def makePair(z):
    return  [(z[i], z[i+1]) for i in range(0,len(z),2) ]

 
N,  = getIntList()

s = input()
s = list(s)
for d in range(2,N+1):
    if N %d !=0: continue
    dprint(d)
     
    x = s[ : d]
    x.reverse()
    s[ : d] = x
    dprint(s)
print(''.join(s) )