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.
#ls.sort(key=lambda x: (x[0], x[1]))
#n,k= map(int, input().split())
# ls=list(map(int,input().split()))
# for i in range(m):
import sys
import math
#print(s[i],end="")
#n=int(sys.stdin.readline())
from collections import Counter
#n,k= map(int, input().split())
#arr= list(map(int, input().split()))
def divs(n, start=2):
    divisors = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if n % i == 0:
            if n / i == i:
                divisors.append(i)
            else:
                divisors.extend([i, n // i])
    return divisors


n=int(input())
s=input()
t=divs(n)
#print(t)
t.sort()
ans=""
for i in t:
    s=s[:i][::-1]+s[i:]
    #print(s)
print(s[::-1])
#arr=list(map(int,input().split()))
