Today, Osama gave Fadi an integer X, and Fadi was wondering about the minimum possible value of max(a, b) such that LCM(a, b) equals X. Both a and b should be positive integers.

LCM(a, b) is the smallest positive integer that is divisible by both a and b. For example, LCM(6, 8) = 24, LCM(4, 12) = 12, LCM(2, 3) = 6.

Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair?

Input

The first and only line contains an integer X (1 ≤ X ≤ 10^{12}).

Output

Print two positive integers, a and b, such that the value of max(a, b) is minimum possible and LCM(a, b) equals X. If there are several possible such pairs, you can print any.

Examples

Input


2


Output


1 2


Input


6


Output


2 3


Input


4


Output


1 4


Input


1


Output


1 1
from sys import stdin
import math


def mp(l):
    # Multiply elements one by one
    result = 1
    for x in l:
        result = result * x
    return result


def prime_factors(n):
    l = []
    while n % 2 == 0:
        l.append(2)
        n = n // 2
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        while n % i == 0:
            l.append(i)
            n = n // i

    if n > 2:
        l.append(n)
    return l


n = int(stdin.readline().rstrip())
if n == 1:
    print(1, 1)
else:

    l = prime_factors(n)
    if l[0]==l[-1]:
        print(1, n)
    else:
        d = []
        s = list(set(l))
        for i in range(len(s)):
            d.append(s[i]**l.count(s[i]))
        d.sort()

        r = [set()]
        for i in range(len(d)):
            for j in range(len(r)):
                z = r[j].copy()
                z.add(d[i])
                r.append(z)
        a = 999999999999999999999999
        b = 0
        diff = a-b
        for i in range(1, len(r)-1):
            n1 = mp(list(r[i]))
            n2 = n//n1
            if abs(n1 - n2) < diff:
                a = max(n1,n2)
                b = min(n1,n2)
                diff = abs(n1-n2)
        print(b, a)