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.
def factors(n):
    fact = []
    for i in xrange(1,int(n**0.5)+1):
        if n%i == 0:
            if i == n/i:
                fact.append(i)
            else:
                fact.append(i)
                fact.append(n/i)
    fact.append(0)
    fact.sort(reverse = True)
    return fact

n = input()
s = raw_input()
arr = factors(n)
ans = ['' for i in xrange(n)]
i = 0; j = n-1
flag = 1
pos = n-1
for k in xrange(len(arr)-1):
    ind = arr[k]-arr[k+1]
    if flag == 1:
        while ind > 0:
            ans[i] = s[pos]
            pos -=1
            ind -= 1
            i += 1
    else:
        while ind > 0:
            ans[j] = s[pos]
            pos -= 1
            ind -= 1
            j -= 1
    flag = 1-flag
print ''.join(ans)  
