Problem: You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.

Divisor of n is any such natural number, that n can be divided by it without remainder.

Input

The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).

Output

If n has less than k divisors, output -1.

Otherwise, output the k-th smallest divisor of n.

Examples

Input

4 2


Output

2


Input

5 3


Output

-1


Input

12 5


Output

6

Note

In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.

In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.

Solution (in PYTHON3):

Solution:
import sys

input = sys.stdin.readline


def I(): return input().strip()


def II(): return int(input().strip())


def LI(): return [*map(int, input().strip().split())]


import copy
import re
import string, math, time, functools, random, fractions
from heapq import heappush, heappop, heapify
from bisect import bisect_left, bisect_right
from collections import deque, defaultdict, Counter, OrderedDict
from itertools import permutations, combinations, groupby, count, filterfalse
from operator import itemgetter

for _ in range(1):
    n,k = LI()
    l = []
    import math



    def printDivisors(n):
        list = []
        for i in range(1, int(math.sqrt(n) + 1)):

            if (n % i == 0):


                if (n / i == i):
                    list.append(i)
                else:
                    list.append(i)
                    list.append(int(n / i))
        return list


    z = printDivisors(n)
    z.sort()

    x = len(z)

    if k > x:
        print(-1)
    else:
        print(z[k - 1])