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
n=input()
from functools import reduce
from bisect import bisect_left
import fractions
l=list(reduce(list.__add__,([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
l.sort()
if len(l)==2:
    print 1,n
else:
    a=bisect_left(l,int(n**0.5)+1)-1
    l1=[n]
    for i in range(a,0,-1):
        if max(l[i],n/l[i])%min(l[i],n/l[i])<>0:
            if fractions.gcd(l[i],n/l[i])==1:
                l1.append(max(l[i],n/l[i]))
    print min(l1),n/min(l1)
