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.
import math

def inp(): return input()

def inpv(): return [int(x) for x in raw_input().split()]

def inps(): return raw_input()

import math 

# method to print the divisors
def printDivisors(n, s) :
	
	# Note that this loop runs till square root
	i = 1
	div = []
	while i <= math.sqrt(n):
		
		if (n % i == 0) :
			
			# If divisors are equal, print only one
			if (n / i == i) :
				div.append(i)
			else :
				# Otherwise print both
				div.append(i)
				div.append(n/i)
		i = i + 1
	div.sort()
	#s = "".join(reversed(s))

	for rb in div:
	#	print rb
		t = s[:rb][::-1]
		s = t + s[rb:]
	#	print s
	print s	
	
def Hills():
	n = inp()
	s = inps()
	printDivisors(n,s)

Hills()