762_A. k-th divisor
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.
```python
#
# Yet I'm feeling like
# 	There is no better place than right by your side
# 		I had a little taste
# 			And I'll only spoil the party anyway
# 				'Cause all the girls are looking fine
# 					But you're the only one on my mind


import sys
# import re
# inf = float("inf")
# sys.setrecursionlimit(1000000)

# abc='abcdefghijklmnopqrstuvwxyz'
# abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod,MOD=1000000007,998244353
# vow=['a','e','i','o','u']
# dx,dy=[-1,1,0,0],[0,0,1,-1]

# from collections import deque, Counter, OrderedDict,defaultdict
# from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
# from math import ceil,floor,log,sqrt,factorial,pow,pi,gcd,log10,atan,tan
# from bisect import bisect_left,bisect_right
# import numpy as np


def get_array(): return list(map(int , sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()


def all_factors(n):
    """returns a sorted list of all distinct factors of n"""
    small, large = [], []
    for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
        if not n % i:
            small.append(i)
            large.append(n // i)
    if small[-1] == large[-1]:
        large.pop()
    large.reverse()
    small.extend(large)
    return small

n,k=get_ints()
result=all_factors(n)
if k>len(result):
    print(-1)
else:
    print(result[k-1])
```