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:
from sys import stdin
import random
import math


def power(x, y, p):
    res = 1
    x = x % p
    while y > 0:
        if y & 1:
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res


def miillerTest(d, n):
    a = 2 + random.randint(1, n - 4)
    x = power(a, d, n)

    if x == 1 or x == n - 1:
        return True
    while d != n - 1:
        x = (x * x) % n
        d *= 2

        if x == 1:
            return False
        if x == n - 1:
            return True
    return False


def isPrime(n, k):
    if n <= 1 or n == 4:
        return False
    if n <= 3:
        return True

    d = n - 1
    while d % 2 == 0:
        d //= 2
    for i in range(k):
        if not miillerTest(d, n):
            return False
    return True


n, m = map(int, stdin.readline().rstrip().split(" "))

if isPrime(n, 4):
    if m == 1:
        print(1)
    elif m == 2:
        print(n)
    else:
        print(-1)
else:
    if n == 1:
        l = [1]
    else:
        l = [1,n]
    for i in range(2, int(math.sqrt(n))+1):
        if n%i==0:
            l.append(i)
            if n//i !=i:
                l.append(n//i)

    l.sort()
    if m > len(l):
        print(-1)
    else:
        print(l[m-1])