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 math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools

sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)

def divisions1(n):
    sq = int(math.sqrt(n)+1)
    d = collections.defaultdict(int)
    while n % 2 == 0:
        n //= 2
        d[2] += 1
    i = 3
    while n > 1 and sq >= i:
        if n % i == 0:
            n //= i
            d[i] += 1
        else:
            i += 2

    if n > 1:
        d[n] += 1

    r = [1]
    for k, v in d.items():
        for c in r[:]:
            for i in range(1,v+1):
                r.append(c*(k**i))

    return sorted(r)

def divisions(n):
    sq = int(math.sqrt(n)+1)
    ss = 30030
    a = [None] * (ss)
    a[0] = 1
    t = 1
    for i in range(2,ss):
        if not a[i] is None:
            continue
        if t * i > ss:
            break
        t *= i
        for j in range(i*2,ss,i):
            a[j] = 1
    a = [i for i in range(t) if a[i] is None]
    d = collections.defaultdict(int)
    for i in a[1:]:
        while n % i == 0:
            n //= i
            d[i] += 1

    for j in range(1,sq//t+2):
        if j > n:
            break
        for ii in a:
            i = t*j+ii
            while n % i == 0:
                n //= i
                d[i] += 1

    if n > 1:
        d[n] += 1

    r = [1]
    for k, v in d.items():
        for c in r[:]:
            for i in range(1,v+1):
                r.append(c*(k**i))

    return sorted(r)

def main():
    n,k = LI()

    d = sorted(divisions(n))
    if len(d) < k:
        return -1

    return d[k-1]


print(main())